Electricity

The second statement defines "volt" correctly.

Explanation: One Volt is the electric potential energy per unit charge, and is measured in joules per coulomb.

p.d. stands for potential difference. Potential difference between two points in a circuit is the work done in moving unit charge (i.e. one coulomb) from one point to the other.

Voltmeter is a device which is used to measure potential difference between two point in an electric ciruit.

Electric Potential at a point being 1 volt means that 1 joule of work is done in moving 1 unit positive charge from infinity to that point

1 Joule of work is done when 1 coulomb of charge moves against a potential difference of 1 volt.

As potential difference =

The SI unit of potential difference is volt, which is also known as Joule/Coulomb

Given, charge = 2 C

Potential difference between two points = 12V

Work done (W) =?

We know that;

V = W/Q

Or, W = V × Q

Or, W = 12 V × 2 C = 24 J

The SI unit of electric charge is coulomb (C).

One coulomb is that amount of electric charge that repels an equal and similar charge with a force of 9 × 10⁹ N when placed in vacuum at a distance of one metre from it.

(a) Potential difference is measured in volts by using a voltmeter placed in parallel across a component.

(b) Copper is a good conductor plastic is an insulator

The law stating that the direct current flowing in a conductor is directly proportional to the potential difference between its ends. It is usually formulated as V = IR. This law is called Ohm's law.

The ohm (symbol: Ω) is the SI derived unit of electrical resistance, named after German physicist Georg Simon Ohm.

The ohm (symbol: Ω) is the SI derived unit of electrical resistance.

Substances having infinite resistivity are called insulators Ex: wood.

As we know that

V = IR

So

As Current is directly proportional to potential difference

So Current is also become half

The strength of electric current flowing in a given conductor depends upon the potential difference and resistance of the conductor.

The larger the cross sectional area, the lower the resistance since the electrons have a larger area to flow through.

As we know that

V = IR

As current is directly proportional to potential difference and inversely proportional to the resistance of the conductor. If resistance of a circuit is halved than the current will doubled.

As we know that

V = IR

As we know that

V = IR

V = 2A × 20 ohms = 40 V

As we know that

V = IR

Ohm's law states a relation between potential difference and current.

On the basis of their electrical resistance, all the substance can be divided into three groups they are Good conductors, Resistors and Insulators.

Those substances which have very low electrical resistance are called as good conductors. A good conductor allows the electricity to flow through it easily. Silver is the best conductor of electricity. Copper and aluminum are also good conductors.

Those substances which have comparatively high resistance than conductors are known as resistors. The alloys like nichrome, manganin and constantan, all have quite high resistance, so these are resistors.

Those substances which have infinitely high electrical resistance are called insulators. E.g- rubber and wood.

Those substances which have very low electrical resistance are called as good conductors like mercury, aluminum, iron, metal coin.

Those substances which have comparatively high resistance than conductors are known as resistors. The alloys like nichrome, manganin.

Those substances which have infinitely high electrical resistance are called insulators like rubber, polythene, wood, bakelite, paper, thermocol

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship

I = V R

Where I is the current through the conductor in units of amperes, V is the voltage measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current.

One ohm is defined as the resistance of an object when a current of one ampere flows through a an object with a potential difference of one volt.

The resistance of a conductor is the ratio of potential difference between the ends of a conductor to the current flowing through it, or the property of a conductor due to which it opposes the flow of current through it is called resistance of the conductor.

Potential differences(V) = Current (I) × Resistance(R)

According to Question,

V = 12volt

I = 2.5mA= 2.5x10^-3A

V=IR

∴

The ohm is defined as a resistance between two points of a conductor when a constant potential difference of 1.0 volt, applied to these points, produces in the conductor a current of 1.0 ampere, the conductor not being the seat of any electromotive force.

The resistance of the wire will increase if we made the conductor thinner because p (rho). l . Area is directly proportional to the radius of the wire.

As we know that

V = IR

So, I = V/R

According to the questions

V= I'2R

So, I'= V/2R

I' = I/2

So current become half

Rubber is an insulator and does not allow the passage of electric current through it, so when electricians wear rubber gloves electric current cannot pass through them, so they don't get shock.

As we know that

V = IR

Here I = 6A, R = 40Ω

So V = 6 × 40 = 240V

(i)

(ii) From graph we find that current values corresponding to potential difference of 0.8V, 1.2V and 1.6V respectively are 0.3 A, 0.45A and 0.6A.

From these values we conclude that ration V/I is a constant i.e., V α I. It means that Ohm’s law is being followed.

(iii) The resistance of the wire is equal to the ratio of potential difference applied and the current passing through it.

Resistance is the ratio of potential difference across a component to the current flowing through it, it is measure in ohms.

The plot between voltage and current is called IV characteristics. The voltage is plotted on y- axis. The values of the current for different values of the voltage are shown in the given table

The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table

From these values we conclude that ration V/I is a constant i.e., V α I. It means that Ohm’s law is being followed.

The formula is given by Ohm which states the relation between potential difference, current and resistance.

Potential difference = Current × Resistance

Here, V=240 volt, I= 5 A, R=?
According to Ohms Law:-
V= IR

As we know that

As we know that

Ω

R= 4 ohms

V= 12 V

I= 12/4 =3A

I=Q/t

Q=3 C

Ohm’s law gives a relationship between current and potential difference.

Ohm is the S.I unit of electrical resistance.

The substance having infinitely high electrical is called conductor.

Current is inversely proportional to resistance.

Double

I=2.4A

potential difference = 120V:

R= V/I

R= 120/2.4 =50 ohms

Since resistance remains constant therefore if potential difference = 240V

then, I= V/R = 240/50 = 4.8A

The electrical property of a material whose symbol is “omega is resistance.

(a) The I-V graph of the metallic conductor is a straight line pass through the origin. They obey Ohm’s Law, having a resistance that is independent of current. (Ohmic conductors). The variation of current vs voltage is shown below:

(b) Temperature remains constant in ohms law.

Here I₁ = 0.02 amp, V₁ = 10 volt

V=IR

Again

Here I₂ = 250mA=0.25A, R = 500

V₂=I₂R =0.25×500=125V

Given,

I = 200mA = 200 × 10^-3A

R = 4000 ohms

So, V = 200 × 10^-3A × 4×10³

V= 800 volts

The resistance of the wire will decrease if we made the conductor thicker because R (resistance) = p (rho). l (length) / A (area of cross-section or thickness of wire). Area is directly proportional to the radius of the wire.

As the resistance of a wire is directly proportional to length of the conducting wire. If its length is doubled resistance also doubled.

There are four factors that influence the resistance in a conductor. Thickness (cross sectional area of the wire), length, and temperature. The fourth factor is the conductivity of the material that is used.

Silver is the best conductor of electricity.

Mercury is a better conductor than iron as it has lower resistivity than iron.

Copper and aluminium wires usually used for electricity transmission because copper and aluminium are very good conductors of electricity having very low electric resistivity (or resistance).

Nichrome wire is used as a heating element because it is very stable, even at high temperatures.

Nichrome (NiCr, nickel-chrome, chrome-nickel, etc.) generally refers to any alloy of nickel, chromium, and often iron and/or other elements or substances. Nichrome alloys are typically used in resistance wire. They are also used in some dental restorations (fillings) and in other applications.

Firstly nicrome is an alloy, it has a higher resistivity and consequently a higher resistance.Therefore, it will resist the flow of charges more and lead to development of heat faster.

Secondly alloys like nichrome don't oxidise,i.e burn easily at high temperatures.

The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise readily at high temperatures.

(a) The resistance of a wire depends on its dimensions as well as on the conducting ability of the material from which it is made. A long wire has more resistance than a short one.

(b) A thin wire has more resistance than a thick one.

If temperature of a metal decrease the resistance will also decrease.

The presence of impurities in a metal increase the resistance of the metal.

Resistance is measured in Ohms. The resistance of a wire increases as the length increases, as the temperature increases, and as the cross-sectional area decreases.

The electrical resistivity of a material is also known as its specific electrical resistance. It is a measure of how strongly a material opposes the flow of electric current. A definition of resistivity is the electrical resistance per unit length and per unit of cross-sectional area.

Given:
The length of the wire, l=1m.
Diameter of wire, D = 0.2mm
∴ Radius of Wire,r=0.1mm= 0.1x10^-3m
Mathematically, the resistivity is defined as

Resistance of a metallic wire with resistivity of length l and cross sectional area A is

Metals are typically good conductors because of their ability to donate electrons. Most metal atoms have one or two valence electrons, which means that they want to give these away in order to become more stable. Silver and copper are good conductors of electricity because they have free electrons available for conduction.

According to Question,

Length of Wire= 1km=1000m

Diameter of Wire=0.50mm

∴ Radius of wire= 0.25mm=0.25x10^-3m

The current will flow more easily through thick wire. It is because the resistance of a conductor is inversely proportional to its area of cross - section. If thicker the wire, less is resistance and hence more easily the current flows.

a) Resistance of a conductor is directly propotional to the length of conductor.

b) Resistance of a conductor is inversely proportional to the area of cross section of the conductor.

c) Resistance of a conductor depends upon the nature of the material of the conductor. Resistance of a conductor increases on raising the temperature and decreases on lowering the temperature.

If we connect different type of wire (nichrome, copper etc) in ac ircuit made using a battery, ammeter. We will see that the ammeter shows different reading in each case i.e. resistance differes as nature of material changes.

(a) The resistance of a metal wire will increase as it is directly proportional to length.

(b) The resistance of a metal wire decrease as it is inversely proportional to the area of the wire.

(c) Resistance of a conductor depends upon the nature of the material of the conductor. Resistance of a conductor increases on raising the temperature and decreases on lowering the temperature

(a) The resistance of a metal wire is inversely proportional to the area of the wire, so if the area increase resistance decrease.

(b) If diameter increase than area increase and the resistance decreases.

(i) The resistance of a wire is directly proportional to its length and inversely proportional to its area (or square of inverse of diameter). When the length is tripled then its resistance will also becomes three times

(ii) If the diameter get tripled then its resistance becomes one ninth of the initial resistance.

(iii) If the material is changed to one whose resistivity is three times then again its resistance will be three times

Here length of the wire = 1 m

Resistance of the wire = 23 ohms

Resistivity of the wire =1.84 × 10^-6 ohm meter

As we know that

By definitions the resistivity of a material of a conductor is the resistance offered by a unit length and unit cross-section of the conductor. It is intrinsic property of material.

The expression for the resistivity is

Where m = mass of the electron

N= number density of electron

E = charge on the electron

Ƭ = relaxation time.

Resistivity is commonly represented by the Greek letter ρ (rho). The SI unit of electrical resistivity is the ohm-metre (Ω⋅m).

The following points state the differences between Resistance and Resistivity:

Resistivity of a substance depends on the nature of the substance and its temperature. It does not depend on the length or thickness of the conductor.

We know,

According to Question,
Length of the conductor(In this case wire) = 1m
Diameter of the wire= 0.3mm

Resistance of the conductor(In this case wire)= 26Ω

When the diameter of a wire is doubled, its resistance becomes one-fourth.

Alloy of nickel, chromium, manganese and iron having a resistivity of about 60 times more than that of copper.

If the diameter of a resistance wire is halved, then its resistance becomes four times.

The resistivity of semiconductor like silicon and germanium is in between those of conductor and insulator.

When the area of cross-section of a conductor is doubled, its resistance becomes

The resistivity of copper metal depends on temperature.

If the area of cross-section of a resistance wire is halved, then its resistance becomes twice.

We known that

Now if we stretch the same wire in such a way that its length get double. At the same time its area of cross-section decrease to half. So that

(a) Material Q with resistivity 2.63×10^-8 ohm-m can be used for making electric wires because it has very low resistivity.

(b) Material R with resistivity 1.0 ×10¹⁵ ohm-m can be used for making handle of soldering iron because it has very high resistivity.

(c) Material P with resistivity 2.3 × 10³ ohm-m can be used for making solar cell because it is a semiconductor.

(a) Good conductor = C (10×10^-8 ohm-m)

(b) Resistor= A (110×10^-8 ohm-m)

(c) Insulator= B (1×10¹⁰ ohm-m)

(d) Semiconductor= D (2.3 ×10³ ohm-m)

(a) E is the best conductor of electricity due to its least electrical resistivity.

(b) C, because its resistivity is lesser than that of A.

(c) B, because it has highest electrical resistivity

(d) C and E, because of their low electrical resistivities.

Those substance which allow electric current to pass through them easily is called conductors. Examples. Iron, copper, zinc e.t.c

Those substance which do not allow electric current to pass through them is called Insulators. Examples. Mica, Air, Paper, e.t.c

Conductor allow electric current to pass through them, so here conductors are Silver, Copper, Aluminum, Nichrome, Graphite, Mercury, Manganin and insulator do not allow current to pass through them, so here insulators are Sulphur, Cotton, Air, Paper, Porcelain, Mica, Bakelite, Polythene

Electronic Potentiol or Potential is defined as the work done in moving a unit positive charge from infinity to a particular point. It is denoted by V and its unit is "Volt".

Electric potential difference is known as voltage, which is equal to the work done per unit charge to move the charge between two points against static electric field.

Here charged moved = 4 C

Potential difference = V₂ – V₁ =230V – 220V = 10 V

So as we know that

Work done = Potential difference × Charge moved

= 10 × 4 = 40 joule

So, the work done in moving a charge of 4 coulombs from a point at 220 volts to another point at 230 volts is 40 Joule.

Voltmeter is an apparatus used to measure the potential difference or electric potential difference between two points in an electric circuit.

Here potential difference = 12 V

Charge moved = 1 C

Work done = Potential difference × Charge moved

= 12 × 1 = 12 joule

So 12 J energy is transferred by a 12 V power supply to each coulomb of charge which it move around a circuit.

When resistors are joined from end to end, it is called in series. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system.

According to the law of combination of series connection

We know that

When resistors are joined in parallel, the reciprocal of total resistance of the system is equal to the sum of reciprocal of the resistance of resistors.

We know that in parallel combination, the reciprocal of total resistance is

According to the question the resultant resistance is less than the individual resistances, so the resistances should be connected in parallel not in series.

The resultant resistance will be less than either of the individual resistance in parallel connection.

Here, R₁=2 ohm, R₂=6 ohm

When the resistors are arranged in parallel combination we get,

Again, when the resistors are arranged in series combination we get

R = R₁ + R₂ = 2 + 6 = 8ohm

When two 4 ohm resistors are connected in parallel then the equivalent resistance will be

1/R_equivalent = 1/4 + 1/4 = 2/4 = 1/2

Or R = 2 ohms

When two 4 ohm resistors are connected in series then the equivalent resistance will be

R_equivalent = 4 ohm + 4 ohm = 8 ohm

Equivalent resistance in arrangement A is 10 ohm as this is in series.

And in arrangement B both 10 ohm and 1000 ohm are in parallel.

So, combined resistance of arrangement B is caculated as follows:

1/R = 1/10 + 1/1000 = (100+1)/1000

R = 1000/101 = 9.9 ohm

So, arrangement in B has lower combined resistance.

As the wire is cut into two equal pieces. So resistance of each part is R/2.

The two parts are joined in parallel. So the equivalent resistance will be

1/R_equivalent = 2/R + 2/R

1/R_equivalent= 4/R

R_equivalent = R/4

(a) Here both resistance i.e. R₁ = 500 ohm, R₂ = 1000 ohm are connected in series

So equivalent resistance R = R₁ + R₂ = 500 + 1000 = 1500ohm.

(b) Here both resistance i.e. R₁=2ohm, R₂=2ohm are connected in parallel

So equivalent resistance

1/R=1/R₁+1/R₂

1/R=1/2+1/2

R=1ohm

(c) Here resistance R₁=4ohm and R₂=4ohm are in parallel and R₃=3ohm are in series

According to the given figure,

1/R=1/R₁+1/R₂

1/R=1/4+1/4

R=2ohm

Total resistance of the given circuit =R+R₃

=2+3=5ohm

Let R₁=6Ὠ, R₂=4Ὠ and V=24V

As the two resistances are connected in parallel. So the current across R₁ will be

And the Current across R₂ will be

In a circuit, resistors are connected end-to-end are said to be in series, if the same current

exists in all of them through a single path. When resistances are connected in series, their equivalent resistance is equal to the sum of the individual resistances.

R = R₁ + R₂ + R₃ + - - - - -

When resistances are connected in parallel, the reciprocal of their equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

1/R = 1/R₁ + 1/R₂ + 1/ R₃ + - - - - -

The resultant resistance is less than either of the individual resistances.

Here a battery of 9 V is connected in series with resistors of R₁=0.2ohm, R₂=0.4ohm, R₃=0.3ohm, R₄=0.5ohm, R₅=12ohm,

So the resultant resistance = R₁ + R₂ + R₃ + R₄ + R₅

R = 0.2+0.4+0.3+0.5+12=13.4ohm

As we know that

V= IR

Thus the current flow through 12ohm resistance will be equal to the current flowing across the whole circuit as in case of Series Connection same current flows through each resistance

∴ I = V/R

I = 9/13.4

I = 0.67amp.

(a) Here resistance are connected in series

Total resistance of the circuit = R₁+R₂=20+4=24Ω(ohm)

(b) According to ohm’s law.

V=IR

Therefore,

6V=I x 24Ω(ohm)

I=6 V /24 ohm = 0.25A(Ampere)

(c) Potential difference across the electric bulb

V₁= IR₁=0.25 X 20=5V

(d) Potential difference across the resistance wire

V₂= IR₂=0.25X4=1V

(i) According to the figure.

R₁ and R₂ are connected in parallel.

Let I be the current following in the circuit, which is equal to 1A. After passing through the resistance R₃

Current I is divided into two part say I₁ and I₂

(ii) Potential difference across AB = IR₃ = 1 x 5 = 5V

As R₁ and R₂ are connected in parallel

So the equivalent resistance 1/R_equivalent = 1/R₁ +1/R₂

1/R_equivalent= 1/10 + 1/15

R_equivalent= 6 ohm

Total resistance in the circuit

R_total= = 5+6 = 11 ohm

Potential difference across AC = IR = 1x11 = 11V

The resistance of 6 ohm and 3 ohm are connected in series. Therefore, their net resistance can be calculated as

R= R₁+R₂

Here, R₁= 6ohm

R₂= 3ohm

so R= 6ohm +3ohm= 9 ohm

The current through 6 ohm resistor= current through line 1 = I= V/R =4/9=0.44V

(ii) The resistance of 12ohm and 3 ohm are connected in series. Therefore, their net resistance can be calculated as

R= R₁+R₂

Here, R₁= 12ohm

R₂= 3ohm

so R= 12ohm +3ohm= 15 ohm

The current through them = I= V/R =4/15

Potential difference across 12 ohm resistor= 4/15 ×12 =3.2

(i) For obtaining minimum current, the two resistors should be connected in parallel.

(ii) For obtaining maximum current, the two resistors should be connected in series.

(a) R₁= 5 ohm, R₂= 10 ohm and V = 6 V

For parallel combination

R= 10/3 ohm

Total current in the circuit

For series combination

R=R₁+R₂

R=5+10=15 ohm

Total current in the circuit.

(b) So the current in each case is 1.8 A and 0.4A in parallel and series circuit respectively.

(i) As shown in the figure, the resistor R₂ and R₃ are connected in parallel. Their total resistance is given by

Here R2= 3 ohm and R3=6 ohm

So

R =2 ohm

The resistance R₁ is in series with the equivalent resistance which are parallel.

Total resistance of the circuit = 2+4ohms = 6ohms

(ii) Total current flowing through the circuit=Potential difference/total resistance

Here p.d =12V, and total resistance = 6 ohms

I = 12/6 = 2amps

(iii) The potential difference across R₁.

V = IR₁=2×4 =8V

Given I= 1 A (Across 5 ohm)

R= 5 ohm

The potential drop across AB, V = I × R

or V= 5 ohm × 1A = 5V

In a parallel circuit, the potential difference across the ends of all resistors remains the same. Therefore the potential difference across 4 ohm and 10 ohm will be 5 V

The current flowing through the 4 ohm resistance

= V/R = 5/4 = 1.25 A

Current through 10 ohm resistor = V/R = 5/10 = 0.5 A

We Know according to Ohms Law that:-

V= IR
Where V= Voltage/Potential Difference
I= Current
R=Resistance

According to Question :-V= 220V
I= 5A
∴ Total Resistance of the Combination is :-

Suppose x resistors should be connected in parallel to draw a current of 5A

for parallel combination:-

According to the question,

x=176/44=4 resistor

Hence 4 resistors each of 176Ω should be connected in parallel so as to draw the current of 5A from a 220 volt supply line.

(a) When only one coil, A is used

V= IR

220= 24I

I=9.2

(b) When coils A and B are used in series.

Total resistance R= R_A +R_B = 24+24 = 48ohms

I = V/R = 220/48

=4.58amps

(c) When coils A and B are used in parallel.

R=12ohms

I = V/R = 220/12

=18.33amps

(a) when three resistors are connected in parallel, the net resistance can be obtained as followed

The resistance of 30 ohm, 20 ohm and 60 ohm are connected in parallel. therefore, the net resistance R will be

----------------------- (1)

The resistance of 10 ohm and 60 ohm are connected in parallel. Therefore, the net resistance R will be

Now R₁ and R₂ are connected in series

So total resistance in the circuit is R = 10+8 =18 ohm

(b) Total current flowing in the circuit

(a) Let I₁, I₂ and I₃ be the current flowing through the resistors of 5 ohm, 100 ohm and 30 ohm, respectively.

According to ohm's law V=IR

Here V= 12 V and R = 5 ohm

I = V/R

Current through R₁ = V/R₁ = 12/5 = 2.4 A

Current through R₂ = V/R₂ = 12/10 = 1.2 A

Current through R₃ = V/R₂ = 12/30 = 0.4 A

(b) Total current in the circuit =Current through R₁ +Current through R₂ +Current through R₃

= 2.4 + 1.2 + 0.4 = 4 A

(c) Total resistance in the circuit = R

R= 3 ohm

(a) When two resistors are connected in series, their resultant resistance is given by

R= R₁+R₂

Here R₁ = 6 ohm

R₂= 2 ohm

Combined resistance, R = R₁ + R₂ = 6+2 = 8 ohm

(b) We know that the current, I= V/R

Here R = 8 ohm

V= 4V

I = 4/8 = 0.5 amp

(c) Potential difference across 6ohm resistor V= I x R₁ = 0.5 x 6 = 3 V

(a) ∵ the 3 Ω and 6 Ω resistance are in parallel, hence the eq. resistance =

(b) I = V/R_eq = 4/2 = 2A

(c) Current Flowing in 3 Ω resistor =

When two resistors are connected in series, their resultant resistance is given by

R= R₁+R₂

Here R₁ = 2 ohm

R₂= 3 ohm

Combined resistance, R = R₁ + R₂ = 2+3 = 5 ohm

(b) We know that the current, I= V/R

Here R = 5 ohm

V= 10V

I = 10/5 = 2 amp

(c) Potential difference across 2ohm resistor V= I x R₁= 2×2=4V

(d) Potential difference across 3ohm resistor V= I x R₂= 2×3=6V

A

The resistors of 6 ohm and 3 ohm are connected in parallel. Therefore, their combined resistance can be calculated as

R= 2 ohms

(b) The p.d. across the combined resistance, V= IR

Here I = 6A and combined resistance = 2 ohms.

So, V= 6 x 2 = 12 V

(c) In parallel connection potential difference remains constant, so Potential difference across 3ohm resistor =12 V

(d) The current in the 3Ω resistor

I= V/R₁

I= 12/3 =4A

(e) Current flowing through the 6 ohm resistor = V/R₂ = 12/6 = 2 A

A

(b) Since two resistors of 20 ohm are connected in series, the effective resistance will be

R= R₁ +R₂

Here, R₁ = 20 ohm

R₂ = 20 ohm

Therefore the effective resistance of the two resistors = 20+20 =40 ohm

(c) Current flowing through the circuit = I = V/R = 5/40 = 0.125 amps

(d) p.d. across each resistance = I x R = 0.125 x 20 = 2.5 V

(a) In the given figure resistors are connected in parallel.

(b) In a parallel arrangement, the voltage remains the same across each resistor i.e 6V

(c) Due to lower resistance 2 ohm have bigger share of current.

(d) The resistors of 2 ohm and 3 ohm are connected in parallel. Therefore, their combined resistance can be calculated as

R= 1.2 ohm

(e) Current flowing through battery, I=V/R=6/1.2=5amps

The coils of the resistors 4 ohm and 2 ohm are connected in parallel. Therefore

R=4/3 ohm

Total current =3 A (given)

Potential difference V= 3×4/3=4V

Now current through 2Ω coil= V/2 =4/2 = 2A

Let the current in the circuit be I amperes and the battery be of strength V volts. Let the combined resistance of the three resistors be R ohms.

Therefore, according to Ohm's law, we have

V=IR ---- (i)

We know that when resistors are connected in series, the current with a battery of V volts.

V₁=I×R₁ ------- (ii)

V₂ =I×R₂ ------ (iii)

Let the potential difference across R₁ is V₁ and the potential difference across R₂ is V₂

V=V₁+V₂--------- (iv)

From equation (i),(ii), (iii) and (iv) we get.

IR= I×R₁+ I×R₂

R= R₁+ R₂

(i) Current through 5 ohm resistor

I= V/R

I = 10/5 = 2A

(ii) Two resistances are connected in series. So same current flow through the circuit,

So current flowing through R =2A

(iii) As we know that according to ohm's law

V=IR

So R= V/I = 6/2 = 3 ohm

(iv) Total V in the given circuit = V₁+V₂ = 10+6= 16 V

Let the current be I amperes and the battery be of strength V volts. Let the combined resistance of the three resistors be R ohms.

Therefore, according to Ohm's law, we have

V= IR -----(i)

We know that when the resistors are connected in series, the current is the same in all the resistors. Therefore

V= V₁ +V₂+ V₃ ------------------ (ii)

Let the current flowing through the whole circuit is I, and the equivalent resistance be R.

According to ohm's law

V= IR

V₁= IR₁ -------- (iii)

V₂= IR₂-------- (iv)

V₃= IR₃ ------ (v)

By using eq (i),(ii),(iii) ,(iv) and (v) we get

IR= IR₁+ IR₂+ IR₃

R= R₁+ R₂+ R₃

(i) Here R₁= 5 ohm, R₂= 10 ohm and R₃= 30 ohm

The value of current through each resistor is different as they are arranged in parallel connection.

So the current through 5 ohm resistor, I₁ = V/R = 6/5 = 1.2A

So the current through 10 ohm resistor, I₂ = V/R = 6/10 = 0.6A

So the current through 30 ohm resistor, I₃ = V/R = 6/30 = 0.2A

(ii) The total current in the circuit = I= I₁+ I₂+ I₃

1.2A + 0.6A + 0.2A = 2A

(iii) Total resistance in the circuit = R, As the circuit is in parallel connection, so

R= 3 ohm

Let the individual resistance of the two resistors be R₁ and R₂ and their combined resistance be R. Let the total current flowing in the circuit be I and strength of the battery be V volts. Then, from Ohm's law, we have: V= IR... (1)

We know that when resistors are connected in parallel, the potential drop across each resistance is the same. Therefore: =I=I₁ + I₂ = V/R₁ + V/R₂

I = V/(1 /R₁ + 1/R₂) ... (2)

From the equations (1) and (2) we have:

1/R = 1 /R₁ + 1/R₂

(i) Let the total resistance of the circuit =R

As the connections are parallel. So the total resistance will be

1/R=1/R₁+1/R₂

R₂=3+2=5ohms

R₁=5ohms

1/R=1/5+1/5

1/R=2/5

R=2.5ohms

(ii) Current shown by the ammeter A will be equal to the current following in the circuit.

I=V/R=4/(2.5)

=1.6amps

Let the resistance of the three resistors be R₁, R₂ and R₃, respectively. Let their combined resistance be R.

Let the total current flowing in the circuit be I and the strength of the battery be V. Then from Ohm's law, we have: V= IR------- (1)

We know that when the resistors are connected in parallel, the potential drop across each resistance is the same. Therefore:

I = I₁+I₂+I₃

I = V/R₁+ V/R₂+ V/R₃

1/R=1/R₁+1/R₂+ 1/R₃

If two resistances are connected in parallel, then the resultant resistance will be

1/R=1/R₁+1/R₂+ 1/R₃

When switch is open, the upper two resistances are connected in parallel in the circuit.

Effective resistance is 1/Req=1/R+1/R=2/R

Req = R/2

So the current=I=V/(R/2)=0.6A (given)

V/R = 0.3 A

When the switch closes, the third resistance also comes in the circuit.

The effective resistance of the circuit becomes R/3

Hence, Current I = V/(R/3) = 3 (V/R) = 3 x 0.3 = 0.9 A

Here two resistor 6 ohm and 2 ohm are connected in parallel

Now R ohm and 6 ohm are in series

So the combined resistance

When the resistors are in series, then the combined resistance = 35 Ω+15 Ω = 50Ω

Now 50Ω and 40 Ω are connected in parallel, so the combined resistance

the two 6 Ω resistors are in parallel. Hence, the Equivalent Resistor =

Now, 3 Ω and 6 Ω resistors are in series. Hence the Total Equivalent Resistance = 3 + 6 = 9 Ω

Here all the resistance are connected in series. Therefore net resistance = 2Ω+3Ω+1Ω =6Ω

So current in the circuit =

Potential across 3Ω resistor

The current at Y is greater than that at Z due to the arrangement of the circuit.

As V = IR

When the wire is cut into 5 pieces, resistance of each part is R/5

So, equivalent resistance of the 5 pieces when connected in parallel is

R₁ = R/25

So required ratio R: R₁=25:1

(i)

To obtain 9Ω resistance two resistance are connected in parallel,

Let the resultant resistance for parallel circuit=R

1/R=1/6+1/6

1/R=2/6

R=3

Effective resistance=6+3 = 9ohms

(ii) To obtain 4Ω resistance all resistance are connected in parallel

So the effective resistance will be R for each circuit

Or R= 2 Ω

Total resistance = 2 Ω+2 Ω=4Ω

Parallel combinations of the resistors will be written as

1/x + 1/8 = 1/4.8

On solving the above relation,

x = 12 ohms

(i) 6 ohm can be obtained by connecting the 1,2 and 3 ohm in series.

(ii) 6/11 ohm can be obtained by connecting the 1,2 and 3 ohm in parallel connection.

(iii) 1.5 ohm can be obtained by connecting 1 and 2 ohm in series and then by connecting 3 ohm to the resultant resistance i.e 3 ohm in parallel connection

Resistor 2 and 3 ohm are connected in series, 5 ohm resistance is connected parallel to 2 and 3 ohm.

Thus the resultant resistance will be 2.5 ohm

(a) For total Resistance to be 4

we will connect 2 ohm resistance in series with parallel combination of 3 and 6 ohm resistance. When 3 and 6 ohm are in parallel combination then 1/R=1/3+1/6

or R=2

Now this R connected in series with 2 ohm resistance will give equivalent resistance = 2+ 2 = 4 ohm.

For total Resistance to be 1 ohm

All the given resistance should be connected in parallel combination 1/R’= 1/2+1/3+1/6

Or R'=1 ohm

(a) To obtain the highest resistance we must connect the given resistances in series.

Highest resistance R = 4 + 8 + 12 + 24 = 48ohms

(b) To obtain the lowest resistance we must connect the given resistances in parallel.

1/R=1/4+1/8+1/12+1/24

On solving we get, R=2ohms

According to the figure, 20 ohm, 10 ohm and 20 ohm are connected in series so the resultant resistance in the extreme side = 20+10+20 = 50 ohms

Now this 50 ohm is in parallel with 30 ohm, so the resultant resistance become

On solving we get R = 18.75 ohm

Now 18.75 ohm, 10 ohm and 10 ohm are connected in series. So the resultant resistance become

R= 18.75+10+10 = 38.75 ohm.

The greatest possible resistance is the series combination of the resistors.

In series combination of hundred 1 ohm resistors equivalent resistance will be 100 ohm

The least possible resistance is the parallel combination of the resistors.

When resistances are connected in parallel then

1/R =100

R= 1/100 =0.01 ohm

To make equivalent resistance of 250 ohms one may connect:

1) 2 resistors in series Thus, the resultant would be:

100+100= 200 ohms

2) 2 resistors in parallel Thus, the resultant of 2 parallel combination would be:

3) Now connect the combination of the two (series and parallel) in series: Thus, we may obtain a net resistance of: 200+50 = 250 ohms

When four 16 ohm resistances is connected in parallel combination then the equivalent resistance will be

1/R = 1/16 + 1/16 + 1/16 + 1/16 = 4/16

R = 4 ohm

When four such combinations are connected in series, than the total resistance = 4+4+4+4 = 16 ohm.

It is given that the two lamps are identical. So, they will have equal resistance.

The ammeter A₁ reads 0.5 A current. This current will now get distributed equally in the two branches as resistances are equal.

Hence, the ammeters A₂ and A₄ will read 0.25 A each.

Now, A₃ and A₅ are connected in series with A₂ and A₄, so they will also read 0.25 A.

No, When appliances are connected in a parallel arrangement, each of them can be put on and off independently. This is a feature that is essential in a house's wiring.

When bulbs are connected in a series arrangement, if one of them can be put off all bulbs are switched off.

When bulbs are connected in a parallel arrangement, each of them can be put on and off independently.

(a) For decorating a hotel building from outside we use series connection.

(b) For lighting inside the rooms of the hotel we use parallel connection.

In series arrangement if one electrical appliance stops working due to some defect, then all other appliance also stop working as the whole circuit is broken.So we should not used series arrangement for connecting domestic electrical appliances in a circuit.

Due to the following advantages different electrical appliances in a domestic circuit are connected in parallel:

(i) Each electrical appliance gets the same voltage as that of the power supply line.

(ii) When appliances are connected in a parallel arrangement, each of them can be put on and off independently.

(iii) If one electrical appliance stops working due to some defect, then all other appliances keep working properly.

(a) Parallel circuit would have the highest voltage across each bulb.

(b) In Parallel circuit the bulbs will be brighter.

(c) In series circuit, if one bulb blows out, all others will stop glowing

(d) Series circuit would have less current in it.

(a) In circuit (ii) the lamps are dimmest as it is connected in series arrangement.

(b) In circuit (iii) the lamps are equal brightness in circuit (i) as both are connected in parallel.

(c) Circuit (ii) give the maximum light.

If we were going to connect two light bulbs to one battery, we would use a parallel arrangement, so the current will flow through them equally. Series arrangement take more current from battery.

When appliances are connected in a parallel arrangement, each of them can be put on and off independently. This is a feature that is essential in a house's wiring.

In series circuit, if one bulb blows out, all others will stop glowing.

The circuit arrangement is in series, so if one bulb blows out, all others will stop glowing.

The lamps in a household circuit are connected in parallel because in a parallel arrangement, each of them can be put on and off independently.

When S₂ open and S₁ closed A and B are lit.

When lamps are connected in parallel the brightness of the lamps increases.

(a) Here A and B are connected in series so the brightness of the A and B will same while C are connected in parallel so the brightness of the bulb C is maximum.

(b) Their will be no effect on A and B.

(c) A and B are connected in series so if B burns than A also not glow but their will be no effect on C, as C is connected in parallel.

The brightness of two lamps which are arranged in parallel is maximum than the the brightness of two lamps arranged in series.

(a) In series circuit, if one bulb blows out, all others will stop glowing.

(b) When lamps are connected in a parallel arrangement, each of them is independently.

(a) To make the light brighter we turn the switch towards right side.

(b) To make the light dimmer we turn the switch towards left side.

The difference in electric potential between two points is known as potential difference. The potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit charge from one point to the other point.

One volt is defined as the work done by one joule to move a charge of one coulomb from one place to another.

As we know that

Here, work done =250 Joule

Charge move =20 C

Voltmeter is an apparatus used to measure the potential difference or electric potential difference between two points in an electric circuit.

The voltmeter is always connected in parallel across the two points where the potential difference is to be measured.

A voltage has a high resistance so that it takes a negligible current from the circuit. The term volt gives rise to the word voltage. Voltage is the other name for potential difference

Two factors on which the electrical energy consumed by an electrical appliance depends on.

1). Power rating of the appliance.

2). Time for which the appliance is used.

As we know that

P is inversely proportional to resistance, so 60 watt has a higher electrical resistance.

The commercial unit of electric energy is Kilowatt-hour.

Here, V = 220 V, P = 100W

R = ?

We know that

Thus

R = V²/P = 220²/100 = 484 ohm

(i) The SI unit of electric energy is joule.

(ii) The SI unit electric power is watt.

(i) Kilowatt is the S.I unit of Electric power.

(ii) kilowatt-hour is the S.I unit of Electric energy.

Watt is the SI unit of Electric power.

kWh is the commercial unit of electrical energy. It full form is kilowatt-hour.

As we know that

i.e. as power is directly proportional to the square of potential difference. So, the electric power becomes four times its previous value.

The electric lamp consumes energy at the rate of 36 J/s.

Here P = 920W, V = 230V, I = ?

We know that

P = V x I

920 = 230 x I

I = 920/230 = 4amp

One watt is defined as the amount of electrical energy consumed by electrical appliances at the rate of 1 joule per second.

1 watt = 1 volt x 1 ampere

The amount of electrical energy consumed by an electrical appliance of 1 watt power for 1 hour is called watt-hour.

1 watt hour is equal to 3600 joules

Here, current (I) = 5amp, Resistance(R) = 100 ohms and time = 2h

As we know that

Electric energy consumed = P x t = I²Rt

= 5² x 100 x 2

= 5000 Wh

= 5 kwh

We know that

1kwh = 3.6 x 10⁶ J

Therefore, 5kwh = 5 x 3.6 x 10⁶ J = 18 x 10⁶ J

Here, Potential difference (V) = 220V, Current (I) = 0.5amp, Power = ?

We know that

P = VI = 220 X 0.5

P = 110 watt

(i) Here current (I) = 1 A, Resistance (R) = 300 ohm, time(t) = 1h

We know that

P = I²R = 1² x 300 = 300 W

E = P x t = 300 x 1 = 300 Wh

(ii) Here Current (I) = 2 A Resistance(R) = 100 ohm, time (t) = 1h

We know that

P = I²R = 2² x 100 = 400 W

E = P x t = 400 x 1 = 400 Wh

Hence, in case (ii), the electrical energy consumed per hour is more.

Here potential difference(V) = 220V, Power (P) = 2.2kW = 2200W, time(t) = 3h

We know that

Electrical energy consumed = P × t = 2.2 ×3 = 6.6 kWh

We have, P = V x I

I = P/V

I = 2200/220

= 10amp

When two electric bulb of 60W are lighted for 4 hours then

Here, p₁ = 60w

Number, n₁ = 2

Time for daily use, t₁ = 4h

Total electrical consumed everyday, E₁ = n₁×p₁×t₁

= 2×60×4 = 480

Energy consumed in 30 days = 30×0.48 = 14.4Kwh (As 1kW=1000W)

When three electric bulb of 100W are lighted for 5 hours then

Here, p₂ = 100w

Number, n₂ = 3

Time for daily use, t₂ = 5h

Total electrical consumed every day, E₁ = n₂×p₂×t₂

= 3×100×5 = 1500

Energy consumed in 30 days = 30× 1.5 = 45Kwh (As 1kW= 1000 W)

Total electrical consumed in 30 days = 14.4Kwh + 45Kwh = 59.4 kWh

(i) We know that

V = 250V, I = 0.4amp

Power = VI = 250X0.4 = 100watt

(ii) We have,

V = 250V, I = 0.4amp

We know that

P = I²R

100 = 0.4²XR

R = 625ohm

(a) Here, P = 4kw, V = 220v

I = ?

We know that

Power = VI = 220 × I

4000 = 220I

I = 4000/220= 18.18A

(b) Here, P = 4kw, V = 220v

R = ?

We know that,

P = I²R

P = (18.18)²×R

R = 4000/(18.18)²

R = 12.10 Ω

(c) Here, P = 4kw, time = 2hr

Energy consumed in two hour = P × t

= 4 × 2

= 8kw-hr

(d) If 1kwh = Rs 4.6

Here, Energy consumed in two hour = 8kwh

So, total cost = 8 x 4.6 = Rs 36.8

Here current (I) = 5amp, Potential difference(V) = 220 volt, time = 2h,

We have to find

Power (P) = ?

Energy (E) = ?

We know that

P = V×I

= 220×5

= 1100 watt

= 1.1 kW

Energy consumed, E = P × t

= 1.1 × 2

= 2.2 kWh

TV set uses 0.25 kWh energy whereas toaster uses 1.20 kWh energy. So, toaster uses more energy.

(a) Here V = 6 volt, R₁ = 1 ohm, R₂ = 2 ohm

The connection is in series. So Equivalent resistance = R₁+R₂ = 1 + 2 = 3 ohm

We know that

Total current I = V/R = 6/3 = 2A

Current through R₂ = I₂ = I = 2A

Voltage across R₂ = V₂ = I₂R₂ = 2×2 = 4 ohm

Power used in R₂ = I₂V₂ = 2×4 = 8W

(b) Here, V = 4 volt,R₁ = 12 ohm, R₂ = 2 ohm

We know that, as the connection is in parallel so voltage across R₂ = V₂ = V = 4V

Current across R₂ = I₂ = V₂/R₂ = 4/2 = 2A

As power = IV

Power used in R₂ = I₂V₂ = 2×4 = 8W

(a)

(b) As the connection is in parallel.

Current through AB i.e 40W lamp = I₁ = P₁/V = 40/220 A

Total current drawn from the electric supply = 40/220 + 60/220 = 0.45 A

(c)We know that Energy= Power x Time

Total Power to two bulbs together=40W+60W=100W

According to the question both of the bulbs are being operated together at 1 hour

∴ Energy= 100W x 1hr
Energy= 100Whr

We know 1Whr= 3600 Joules
∴100Whr= 3600 X 100
=360000Joules
=360kJoules

(a) Here, V = 230V, I = 10amp

We know that, P = VI

P = 230X10

P = 2300watt = 2300 J/s

(b) The energy transferred in 1 minute. = P x t = 2300 J/s x 60s = 138000 J

Here for heater we have, P = 2kW, t = 4h

So energy consumed by heater, E = P x t = 2x4 = 8kWh

Here for TV, we have, P = 200W = 0.2kW, time 10pm – 6pm = 4h

So energy consumed by TV, E = P x t = 0.2x4 = 0.8kWh

Here for TV, we have, P = 100W = 0.1kW, t = 4h, number of lamp(n) = 3

So energy consumed by three lamps, E = n x P x t = 3x0.1x4 = 1.2kWh

Total energy consumed by heater, TV and three lamps = 8+0.8+1.2 = 10kWh

Cost of 1kWh = Rs. 5.50

Cost of 10kWh = Rs. 5.50 x 10 = Rs. 55

Here I = 13amp, V = 230V

As we know that

Power = VI

= 230 x 13

= 2990W

P = 2.99kW

Here, V = 230V, I = 0.4amp

We know that power is the rate at which electric energy is transferred

P = V x I

= 230 x 0.4

= 92 W = 92 J/s

The rate at which electrical energy is consumed, or the rate of doing electric work is known as electric power.

We know that

P = VI

(i) resistance = V/I = 3/0.5 = 6 Ω

(ii) P = VI = 3 × 0.5 = 1.5 Watt

The amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour is called One kilowatt hour.

1kWh = 3.6 x 10⁶J

Here, Power of heater (P) = 500W = 0.5kW, Time(t) = 20hr

We know that

Energy consumed = P x t = 0.5 X 20

= 10kwh

Total cost = 10xcost per unit

Cost per unit = Rs. 3.9 per unit

Therefore, total cost = 10 x 3.9 = Rs 39

We know that

P = VI

P = 12× 0.5 = 6W

The unit for expressing electric power is watt

The correct wattage for an electric iron used in our homes is 850 W.

Energy consumed = P x t = 2 X 3 = 6kWh

Total cost = 6 x cost per unit

Cost per unit = Rs. 4 per kWh

Therefore, total cost = 6 x 4 = Rs 24

The SI unit of energy is joule.

The commercial unit of energy is kilowatt-hour.

Energy consumed = P x t = 100 ×60 = 6000 joule

As in 1 minute = 60 second.

As P = VI

I = P/V

As P = VI

When they are connected in parallel 293 bulb will glow in the house at the same time.

If the potential difference between the ends of a fixed resistor is halved, the electric power will become one-fourth.

An electric heater will consume more electrical energy as power is inversely proportional to the resistance.

(a) Lamp has the greatest electrical resistance as resistance oppose the flow of current, and the lamp has least amount of current.

(b) According to the given data large amount of current is needed by kettle, so it must be connected to the earthing.

(c) As we know that

P = VI

V = 240V, I = 8.5A

P = 240X8.5 = 2040 W = 2.04 kW

(d) When kettle is connected to 240 V supply, then the P = 2040 W

R = V²/P =

R = 28.23 ohm

Now, when V = 120V, R = 28.23 ohm

I = V/R = 120/28.23 = 4.25A

(a) The meter reading on Sunday is 42919.

(b) The meter reading on Monday is 42935.

(c) Units of electricity used = 42935- 42919 = 16 units.

(d) The cost of 1 unit is Rs 5. So the cost of 16 units = 16×5 = 80 rupees

Here it is given P = 10W, V = 220V, I = 5A

As we know that

P = VI

= 220X5

P = 1100W

Power of one bulb = 10W

So the total no. of bulbs that can be connected = 1100/10 = 110

As we know that

Electric power consumed = V²/R

Here V is given we have to find the value of R

So when the connection is in parallel

The equivalent resistance is

Electric power consumed in parallel connection

When the connection is in Series

The equivalent resistance = R+R = 2R

Electric power consumed in series connection

The ratio of electric power consumed by them

Heat produced, H = I²Rt

As heat produced is directly proportional to the square of current.

Heat produced, H = I²Rt

So, If the current passing through a conductor is doubled then heat produced becomes four times.

The two effects produced by electric current are:

(a) Heating effect of current

(b) Magnetic effect of current

Heating effect of current is utilized in an electric light bulb.

Heating effect of current is utilized in the working of an electric fuse.

The two devices which work on the heating effect of electric current are Electric heater and electric fuse.

Argon and nitrogen are filled in filament type electric light bulbs,

Due to the heating effect of current lots of electric power is consumed by the filament of a bulb

Due to the low resistance of copper the connecting cord of the heater does not glow because negligible heat is produced in it by passing current but the heating element made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current (because of its high resistance).

The heat produced when a current I is passed through a resistor R for time t will be

Heat produced, H = I²Rt

Here,

R = 20ohm, I = 5amp, t = 30s

We know that H = I²Rt

H = 5²x20x30

H = 15000 J

Heat produced by an electric current is given as H = I²Rt, that means

(i) Heat produced is directly proportional to square of current.

(ii) Heat produced is directly proportional to resistance.

(iii) Heat produced is directly proportional to the time for which current flows.

Joule's law of heating states that when a current of I amperes flows in a wire of resistance R ohms for time t seconds then the heat produced in joules and is given by H = I²Rt

The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance to resistance for a given current and (iii) directly proportional to the time for which current flows through the resistors.

Here it is given that R₁ = 40 ohms, R₂ = 60 ohms, V = 220V, t = 30 sec

Here R₁ and R₂ are arranged in series connection.

So the equivalent resistance = R = 40+60 = 100 ohms

According to Ohm's law, we know that

V = IR

I = V/R

I = 220/100 = 2.2amp

As we know that heat produced in joules is given by H = I²Rt

Substituting the values of I, R and t in eq. H = I²RT

H = (2.2)² X 100 X 30

H = 14520 J

As the filament of the bulb is made from tungsten, so when we fill air inside the electric bulb this tungsten filament will burn quickly. So we fill unreactive gas like argon or nitrogen in the electric bulb to prolong the life of filament.

Due to its very high melting point of tungsten it is used for making the filaments of electric bulbs

Due to the low resistance of the connecting wires of the heater negligible heat is produced in them by passing current so it is slightly warm.

(a), Here it is given,

I = 4 amp,

t = 10 min = 10 x 60 = 600sec, H = 2.88x104J

We Know that according to Joule's law of heating

H = I²RT

28800 = 4²xRx600

R = 3ohms

We know that

P = I²xR

= 4²x3

P = 48W

(b) Here we have to find the potential difference, V = ?

We know that

V = IR

V = 4x3

V = 12V

Here Resistance of the coil(R) = 200 ohms, Current (I) = 2.5 amp, time(t) = 1 sec

We know that according to Joule's law of heating

H = I²RT

H = (2.5)²×200×1

H = 1250 J/s

Here Resistance (R) = 8 ohms, Current(I) = 15 amp, time (t) = 1 sec

We know that according to Joule's law of heating

H = I²RT

H = 15²X8X1

H = 1800J/s

Here R = 25ohms, V = 12V, H = ?, t = 60sec

According to Ohm’s law

V = IR

12 = 25XI

I = 0.48amp

We know that according to Joule's law of heating

H = I²RT

H = 0.48²X25X60

H = 345.6J

Here, it is given H = 100J, t = 1sec, R = 4ohms,

We know that according to Joule's law of heating

H = I²RT

100 = I²×4×1

100/4 = I²

I = 5amp

Again we know that

V = IR

V = 5X4

= 20V

All materials offer resistance to the flow of current through them. So some external energy is required to make the current flow. This energy is provided by the battery. Some of this energy gets dissipated as heat energy, so the resistor becomes hot.

Work done in carrying a charge Q through a potential difference V is given as

Also, Q = I t

Using Ohm's law, V = I R

W = I² Rt

This work done in carrying the charge through the wire appears as the heat produced.

i.e. H = V I t = I² Rt.

this energy is dissipated as heat energy.

This law is called Joule law of heating effect.

Here, it is given that P = 12W, V = 12V, t = 60sec

We know that

P = VI

I = P/V = 12/12 = 1A

Again, we know that

V = IR

R = V/I = 12/1 = 12ohm

So according to the Joule’s heating law, we know that

H = I²Rt

H = 1²x12x60

H = 720J

Joule heating is given by

H = i²Rt

H is the heat produced, i is the current through the wire of resistance R for time t. If i is halved i.e i' = i/2

H become 1/4 th

When electric current is supplied to a purely resistive conductor, the energy of electric current is dissipated entirely in the form of heat and as a result, resistor gets heated. The heating of resistor because of dissipation of electrical energy is commonly known as Heating Effect of Electric Current. Some examples are as follows:

When electric energy is supplied to an electric bulb, the filament gets heated because of which it gives light. The heating of electric bulb happens because of heating effect of electric current.

When an electric iron is connected to an electric circuit, the element of electric iron gets heated because of dissipation of electric energy, which heats the electric iron. The element of electric iron is a purely resistive conductor. This happens because of heating effect of electric current.

Pure tungsten has some amazing properties including the highest melting point (3695 K), lowest vapor pressure, and greatest tensile strength out of all the metals. Because of these properties it is the most commonly used material for light bulb filaments.

The heat produced by passing an electric current through a fixed resistor is proportional to the square of magnitude of current.

As we know that, H = V I t = I² Rt.

The current passing through an electric kettle has been doubled. The heat produced will become four times.

As H = I ² Rt

When the circuit current exceeds a specified value due to voltage fluctuations or short-circuiting, the fuse wire gets heated and melts.

An electric heating element is generally made from nichrome and can come in the shape of either a coil, ribbon or wire strip. When electricity is introduced into the heating element, its internal temperature increases and grows red hot as it begins to radiate heat outward

As H = I ² Rt

So, the heat produced in a wire of resistance ‘x’ when a current ‘y’ flows through it in time ‘z’ is given by x × z x y²

In electronics and electrical engineering, a fuse is a type of low resistance resistor that acts as a sacrificial device to provide over current protection, of either the load or source circuit.

In a filament type light bulb, most of the electric power consumed appears as infra-red-rays.

The automated mass manufacturing of incandescent filaments intended for operation at over 2500 K is a significant technical achievement

If the current flowing through a fixed resistor is halved, the heat produced in it will become one-fourth

As H = I ² Rt

(a) The metals and alloy have very low resistivity in the range of 10^-8 ohm meter to 10^-6 ohm meter, so they are good conductor of electricity. While insulators have resistivity of the order of 10¹² to 10¹⁷ ohm meter. Alloys have generally higher resistivity than its constituent metals and consider good for the heating elements. So S is used for heating element of electric iron.

(b) Q has very low resistivity of 1.7 X10^-8ohm-m (it is actually copper). So it is used in connecting wires of electric iron.

(c) R has very high resistivity of 1.0X10¹⁵ohm-m (it is actually rubber). So it is used for covering of connecting wires.

Because of the property of low melting point it behaves differently from the other wires.

This is because of the high resistance offered by filament wire.

Resultant resistance in a series combination is greater than any individual resistance, and resultant resistance in a parallel combination is smaller than any individual resistance.

Since His proportional to R; so more the resistance, more the heat. Therefore, the resistances must be connected in series to obtain more heat per minute.

To find the heat produced by iron in the problem, we need to use an expression for produced power, which depends on the resistance of the iron and the current flowing through the iron.

P_min = VI

360 = 220XI

I = 1.63amp

R = V/I

R = 220/1.63

R = 134.96ohms

In case of maximum heating we get

We know that

P_max = VI

840 = 220XI

I = 3.81amp

R = V/I

R = 220/3.81

R = 57.74ohms

Electric fuse, Room heater, Electric iron are some of the device which have resistor which control the flow of electricity

Potential difference between two points in a circuit is the work done in moving unit charge (i.e. one coulomb) from one point to the other.

The device used to measure the potential difference between two points in volts is known as voltmeter.

Coulomb is the SI unit of electric charge which is equal to the amount of charge transported by a current of one ampere in one second.

Volt is the electrical unit of voltage or potential difference. One Volt is defined as energy consumption of one joule per electric charge of one coulomb.

One coulomb is the amount of charge that passes a point each second when the current is one amp. One coulomb is also equal to the charge contained in 6.24 x 10¹⁸ electrons.

SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second.

The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor.

When we switch on a light electrons or charge are travelled through the wire.

The electric current is constituted by electron in a metallic conductor.

The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery.

Electrons would actually move through the wires in the opposite direction i.e. from negative terminal to positive terminal.

If 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

Therefore; 1 A = 1 C/1 s

SI unit of electric current is ampere (A).

Small quantity of electric current is expressed in milliampere and microampere. Milliampere is written as mA and microampere as μA

1mA (milliampere)= 10⁻³ A

1μA(microampere)=10⁻⁶ A

If the voltmeter is connected in series then the potential diff between two consecutive points will be zero. Thus voltmeter is useless. Ammeter is used to measure the current flowing in a branch.

If the voltmeter is connected in series then the potential diff between two consecutive points will be zero. Thus voltmeter is useless. Ammeter is used to measure the current flowing in a branch. If it is connected in parallel, then the current will be divided between different branches. So voltmeter is connected in parallel and ammeter is connected in series.

(i) This symbol represents variable resistance or rheostat

(ii) This symbol represents a closed switch or a closed plug key.

The flow of electric charge is known as electric current. Here 20 C charge pass in a circuit in one second.

Here current I = 4 A and Time = 10 sec

So charge = Current × time

= 4 × 10 = 40 C

Here charge =20 C and time = 40 sec

(a) Electric current is carried by moving electrons through a conductor. Electric circuit is a continuous and closed path of electric current.

(b) Current is measured in ampere using an ammeter placed in series in a circuit.

A source of electricity such as cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.

As we know that

Q = I × t

Here I= 2 A and T = 1 minute = 60 sec

Thus Q = 120 C

Again as we know

V = W/Q

W = VQ

W = 10 × 120 = 1200 joule

The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor. It is the potential difference between the ends of the wire which make the electric charge (or current) to flow in the wire.

SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second.

An ammeter is an apparatus which is used to measure electric current in a circuit. An ammeter is always connected in series with the circuit in which the current is to be measured.

The equation below shows the relationship between charge, current and time: charge (coulomb, C) = current (ampere, A) × time

Q = I × t

Here I = 0.36 A

Time = 15 minute = 15 × 60 = 900 sec

So Charge = 0.36 × 900 = 324 C

(a) An ammeter has to measure to current flowing through the circuit. Resistance offers an obstruction to the current flow. So, if the resistance of an ammeter is large, the current measured by the ammeter will be quite less as compared to the actual amount of current flowing through the circuit which is undesirable. If ammeter has zero resistance, then it will give the exact value of current. But this is not practically possible because every material has some value of internal resistance which we can't control. For this reason, ammeter must have small resistance.

(b) If current is flowing through the voltmeter, then it is not all flowing through the load, and the potential difference across the load would change when the voltmeter is added and removed. This is unfavourable. Therefore, the voltmeter must have a very high resistance so that current doesn't flow through it.

Electric circuit is represented by drawing circuit diagrams. A diagram which indicates how different components in a circuit have been connected by using the electrical symbols for the components is called a circuit diagram.

Voltmeter is an apparatus used to measure the potential difference or electric potential difference between two points in an electric circuit. A voltmeter has a large resistance.

We know that charge over 1 electron = 1.6 × 10^-19 coulomb

Thus, 1.6 × 10^-19 C of charge = 1 electron

Therefore, 1 C of charge

As we know that

Work done = potential difference x charge

a) W= 12 ×1= 12 j

b) W= 12 ×5 = 60 j

c) W = 12 × 2 x 10 = 240 j

(a) As we know that

(b)

The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor. SI unit of electric current is ampere (A).

One coulomb is nearly equal to 6 x 10¹⁸ electrons. SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second. This means if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

An ammeter is an apparatus which is used to measure electric current in a circuit. It should be connected in series with the circuit.

The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction i.e from negative terminal to positive terminal.

Here

Charge, Q = 10C

Voltage = 10 MV = 10 x 10⁶V

Energy of the charge particle is given by

E = Q V

= 10 x 10 x 10⁶

=10⁸ J

Electric potential difference is known as voltage, which is equal to the work done per unit charge to move the charge between two points against static electric field. It is denoted by ‘V’.

An ammeter has to measure to current flowing through the circuit. Resistance offers an obstruction to the current flow. So, if the resistance of an ammeter is large, the current measured by the ammeter will be quite less as compared to the actual amount of current flowing through the circuit which is undesirable. If ammeter has zero resistance, then it will give the exact value of current. But this is not practically possible because every material has some value of internal resistance which we can't control. For this reason, ammeter must have small resistance.

SI unit of electric current is ampere (A).

Charge = Current × time = 10 × 5 = 50

Charge = Current × time

(a) The lamps are connected in series. In a series circuit, every device must function for the circuit to be complete.

(b) Here the students connected Ammeter in parallel with the lamps. An ammeter is always connected in series.

(c)

Two bulbs are connected in series, so that both bulbs are controlled by same switch and the third bulb is connected in parallel, so that it can be controlled by another switch.

(a) Here Potential difference = 230 V

Current = 8A

Time = 1 Second

As we know that

The flow of electric charge per unit time is known as electric current.

So in each second the amount of charge flowing in the circuit is 8 C.

(b) As we know that the work done is equal to the amount of energy transferred in a circuit.

Electric potential difference is known as voltage, which is equal to the work done per unit charge to move the charge between two points against static electric field.

So,

So, Work done = 230 × 8 = 1840 J

It is also equal to the total energy transferred to the heater in each second

Here

Q = I × t

As we know that

Q = ne

Ne = I × t