Electricity

If a body has positive charge it means that it has less no. of electrons than the no. of protons. Due to this imbalance of the the charges, the body will attract the negative charged body.

In an electrical circuit the current flows from the positive to the negative. The conventional current always flows in the direction opposite to the direction of flow of the electron.

We know that;

Q = ne

Were,

n is an integral multiple can take values 1, 2, 3….

e^- is the charge on the electron.

And,

Thus,

Ammeter is a device that is connected in series so as to measure the current through the circuit. It has a very low resistance so that maximum current passes through it and thus the correct amount of current through the circuit can be measured.

We know that work done in moving a charge Q in a potential V is given by:

Since, Q = 20mC = 2010^-3C

Voltmeter is a high resistive device that is used to measure the voltage across a component. It is always connected in parallel to the component whose voltage is to be measured. It is highly resistive so that the current cannot pass through it and thus the current will pass through the component connected in parallel to it so as to measure the accurate voltage drop across the component.

If a circuit is closed that means electrons can easily flow through the circuit and thus electric current will flow through it.

If a circuit is open that means the electrons will not be able to flow through the circuit which means that there will be no current flowing through it.

Here, in this circuit, the student has kept the ammeter in parallel and the voltmeter in series with the circuit.

But actually, the ammeter should always be connected in series as it has a very low resistance so when we connect the ammeter in series, maximum current will pass through it and thus the correct amount of current through the circuit can be measured.

In the same way, the voltmeter should always be connected in parallel as it is highly resistive due to which the current cannot pass through it and thus the current will pass through the component connected in parallel to it so as to measure the accurate voltage drop across the component.

Electric resistance (R) is defined as the ratio of the voltage applied (V) to the current (I) flowing through the circuit.

It is given by the formula:

Its SI unit is known as ohm (Ω).

We know that by Ohm’s Law:

Now,

We know that:

Thus, from this we can conclude that if the resistance of the circuit is halved then the current gets doubled.

We know that by Ohm’s Law:

Here: V = 75V

and, R = 30Ω

The formula of resistance is:

Where, resistivity

Length of the conductor

A = Area of the conductor

If length of the conductor is halved then,

And, the cross sectional area is doubled,

Now,

Thus, we can conclude that the new resistance of the wire when the area of wire is doublead and length is haved become � the initial resistance.

Resistivity: It is defined as the resistance offered by a conductor having unit length and unit cross-sectional area. It is denoted by (rho).

The SI unit of resistivity in Ωm.

When different resistors are connected in series combination the current passing through them remains the same.

V_eq =V₁ +V₂ + V₃ +… +V_n)

Since, V= IR

Putting the value in the above equation.

V_eq= I(R₁ +R₂ +R₃…+R_n)

I R_eq=I (R₁ +R₂ +R₃…+R_n)

R_eq=(R₁ +R₂ +R₃…+R_n)

Thus, we can say that the current remain constanct in series connection.

When different resistors are connected in parallel combination the voltage drop remains the same.

We do not use series combination for connecting electrical appliances in household circuit because if there will be a short circuit in one of the appliance which led to break in the circuit will cause the working of other appliance connected in the series.

Thus, Parallel combination is used for connectind electrical appliance in the household circuit.

A Mixed combination of resistances means When two more resistance are connected in series and parallel simultaneously.

These type of combination are known as Mixed combination.

Here, the resistances 10Ω and 15Ω are connected in series and also the resistances 20Ω and 5Ω are also connected in series.

Thus, R₁ = 10Ω + 15Ω = 25Ω

R₂ = 20Ω + 5Ω = 25Ω

Now, these two resistances R₁ and R₂ are connected in parallel

Thus the net resistance is 12.5Ω.

When current flows through a conductor, it produces some amount of heat. This phenomenon is known as the heating effect of electric current.

The heat (H) produced during the heating effect of electric current is given by the formula:

Heat, H = I² Rt

Where, I = Current

R = Resistance of conductor

t = time

The SI unit of the heat produced is Joules (J).

The factors on which heating effect in a current conductor depends are:

a) The time for which the current is flowing through the circuit. Greater the time greater will be the heating effect.

b) Resistance of the conductor. More resistance yields more heating effect.

c) The amount of current flowing through the circuit.

Application: This principle of heating effect of current carrying conductor is used in the iron.

The iron heats up as current flows through it and thus we are able to press the clothes easily.

The fuse wire is made up of lead and tin alloy as this alloy has a very low melting point and a very high resistivity. When a large amount of current flows through this alloy, due to the high resistivity heat is produced due to the flow of high current Which make the alloy melts (as it has a very low melting point). Thus,as alloy melts the circuit breaks and the appliances connected to it are saved from any damage due to the high amount of current flowing in the circuit.

We know that the formula of heat is given by:
Heat, H = I² Rt

Where;

t = 30 min = 30×60 sec = 1800sec

I = 15A

R = 500Ω

Thus Total heat produced by the heater is 20.25× 10⁷ J in 30 min due to flow of 15 A current.

The electric bulbs are filled with chemically inactive as Nitrogen or Argon.

As when the current is passed through the bulb, the buld filament made of tungsten glow due to its high resistivity if exposed to air will cause the decomposition of tungsten metal in air which can leds

Tungsten element to melt thus causing circuit to break.

Thus for the long life of the bulb the electric bulb is filled with chemically inactive as nitrogen and argon.

We know that the formula of power is:

Here;

V = 230V

I = 13A

Thus, the maximum power is 2.99kWcan be drawn from the circuit.

We know that the power can be stated as the amount of electric energy consumed per unit time.

On soving For Energy, we get

Energy,

Here;

P = 60W

t = 1s

Thus, the energy in joules is 60J.

Electric fuse is used as a safety device. Whenever, there is a large amount of current flowing through a circuit due to some faults, due to the lower melting point of the alloy, the fuse melts down which breaks the circuit connection.

Thus, electric fuse is an important component of all domestic circuits.

When current flows through a conductor for a unit time, it produces some amount of heat. This phenomenon is known as the heating effect of electric current.

The heat (H) produced during the heating effect of electric current is given by the formula:

Here, I = Current

R = Resistance of conductor

t = time

The SI unit of the heat produced is Joules (J).

The two common materials, used as heating element are:

i. Tungsten wire.

ii. Nichrome Wire (It is made of 80% of Nickle and 20% of chromium.

1kwh = 1000watt×3600seconds

The SI unit of energy is joules (J).

Thus, we conclude that the 1kwH is 3.6 × 10⁶ J

We know that the formula of power is given by:

So, from this equation we can see that power is inversely proportional to the resistance R. which means that lower is the resistance, more will be the power.

Thus we can say that: 100W will have the lower resistance.

Two features of the material used as an element of an electric iron are:

a) It should have low melting point.

b) It should have a very high resistance.

Parallel Connection: The electrical appliances should always be connected in parallel combination because if there is any fault/damage in any of the appliances then in series combination the whole circuit will get open but in parallel combination only that part containing the damaged appliance is open which will have no affect on the working of other appliances in the circuit.

Series Connection: We do not use series combination for connecting electrical appliances in household circuit as whenever there will be a damage/breakage in the circuit of any one appliance of the household then, due to the series connection all other connections will also break.

The SI unit of energy is joules (J).

The graph obtained by plotting the values of V and I is a linear graph. We will observe that as the voltage V increases the value of current I also increases linearly.

This happens because we know that according to the Ohm’s Law:

Yes, it is possible to replace resistors joined in series by an equivalent single register.

Example: Suppose there are three resistors of 1Ω, 2Ω and 3Ω connected in series.

So, if we want to attach only one resistor instead of these three resistors then we and add these resistors ie.

1Ω + 2Ω + 3Ω = 6Ω

Thus, we can place a single equivalent resistance of 6Ω instead of these three resistors.

The formula of resistance is:

Where, resistivity

Length of the conductor

A = Area of the conductor

Here:

R = 10

r = 0.25mm

l = 75m

We know that the area of the wire will be:

Putting the given values in the above formula,

Here: R₁ = 3Ω

R₂ = 3Ω

R₃ = 3Ω

• Parallel combination:

R1, R2 and R3 connected in stair like pattern one above the other.

• Series Combination:

R1, R2 and R3 connected in same line across the potential difference V

• Mixed combination:

R1 is connected in series with the parallel combination fo R2 and R3.

Putting the values of the resistance, we get

Thus, we get

R_s= R₁+ R_p= 3+1.5= 4.5 Ω

Thus, the net resistance of the circuit is 4.5Ω

Given:

V = 3V

R₁ = R₂ = R₃ = R₄ = 10Ω

Here, R₁ and R₂ are connected in series

And, R₃ and R₄ are connected in series.

Now, by Ohm’s Law

Thus, Current drawn by the resistor from the battery is 0.3 A

i. Here, the new resistance:

The resistivity is dependent on the nature of the material.

Hence, there will be no change in the reisitivity of the wire.

ii. Now, by Ohm’s Law

Thus, the current will increase 9 times than the previous current.

• Suppose 3 resistances are R₁, R₂ and R₃. We will connect these resistances in parallel with an ammeter (A), a voltmeter (V), a plug key (K)and a battery of known voltage as shown in figure given below.

• We will close the key K and record the ammeter and voltmeter readings.

• Now we will open the key and switch the position of voltmeter and ammeter as show in the figure given below.

• We will close the switch and note the readings. Similarly, we will switch the position of voltmeter to resistances R₂ and R₃ and observe the readings.

• We will find that readings of ammeter keeps on changing but readings of voltmeter in all cases almost remain same in all cases.

Here,

₂ = 120cm = 1.20m

R₁ = 5.0Ω

R₂ = 2.5Ω

Now,

And

(Here, resistivity and the area A remain same as the wires are similar.)

Now,

l₁ = 2×120×10^-2 m

l₁ = 24×10^-1 m

∴ l₁ = 240 cm

i)

• Disha connected wire A with both the batteries.

• When wire A was connected with 3v battery, Disha found that the wire started heating.

• When the wire A was connected with 12V battery, she found that the wire is emitting a large amount of heat.

• Thus, she concluded that wire A is a nichrome wire as nichrome wire has a very large resistance and it produces a large amount of heat.

ii)

• Disha connected wire B with both the batteries.

• When wire B was connected with 3v battery, did not find anything.

• When the wire B was connected with 12V battery, she found that the wire has started melting and .

• Thus, she concluded that wire B is a fuse wire as fuse wire has a very low melting point and it melts as a large amount of current is made to pass through it.

The equivalent in all the three circuits comes out to be the same so I_I = I_II = I_III. The reason being, the two end resistors of R and R ohms add upto 2R and the subsequently the two resistors of 2R in parallel are equivalent to a single resistor of R ohms and then the process repeats until the arm AB is reached , so, finally the equivalent resistance of all the circuits is same.

The given circuit can be reduced to : -

The equivalent resistance can be calculated as: -

R_eq= + 0.5 = Ω

R_eq = Ω

The net current in the circuit is given as

= A, so,

I_net = 18/11 A.

The circuit can be drawn as shown aside,

The voltage drop across the parallel resistance combination is , so net current in the 4 Ω resistor is = 0.55 A .

Assuming that each interval is of 1 seconds, we can calculate the heat for each interval and just sum it up to find the net heat produced,

Heat, H= I² RT

Where,

I is the current,

R is the resistance,

T is the time taken

For AD,

I=3, R=2 Ω, T=1 sec

Putting the values in the above formula, we get

H_AD=3² . 2 . 1=18 J

For DG,

I=-2 A, R= 2Ω, T=1 sec

Putting the values in the above formula, we get

H_DG=(-2)² . 2 . 1=8 J

For GJ,

I=1 A, R=2 Ω, T=1 sec

Putting the values in the above equation, we get

H_GJ=1² . 2 . 1=2 J

So Total amount of heat generated in 3 sec

H_NET = 18 + 8 + 2 = 28 J

The statement is correct.

Explanation:The statement is true since for the motion of charges through a material requires a high potential and a low potential.Since, there feets are not touching the ground, the electric charges are not able to flow their body.

Thus we conclude that the electric charges require a potential difference to flow flow across a material. This only the region that the bird sitting on the High tension line are not electrocuted.

If wire A has four times the resistance, then it must have the smaller cross-sectional area since resistance and cross-sectional area are inversely proportional. In fact, A must have one-fourth the cross-sectional area of B. Since the cross-sectional area of a circular cross-section is given by the expression:

wire A must have one-half the radius of wire B and therefore one-half the diameter. Put another way, the diameter of wire B is two times greater than the diameter of wire A.

Here in this diagram, say:

R₁ = 4Ω

R₂ = 8Ω

R₃ = 4Ω

R₄ = 8Ω

When we want to find the resistance across points A and B then resistance R₁ and R₂ are connected in series and resistances R₃ and R₄ are also connected in series.

Thus,

R₅ = R₁ + R₂ = 4Ω + 8Ω = 12Ω

And, R₆ = R₃ + R₄ = 4Ω + 8Ω = 12Ω

Now, this R₅ and R₆ are connected to each other in parallel. Thus the equivalent resistance R_AB across points A and B is:

Given:

Mass of the object = 2 kg,

Charge = 10 C,

Potential difference = 20 V,

Calculations:

We know that the Energy transferred which a charge q is moved through a potential difference V is given by

E = q . V

Where,

Q is the charge

V is the potential difference

Also gravitational potential energy stored in an object is given as:

E= m× g× H

Where,

M is the mass of the object,

G is gravitational acceleration,

H is the displacement,

As per the problem both are equal,

m.g.x = q. V

putting the values in tehequation, we get

2× 10× x = 10 . 20

x =

= 10 meters.

The current when it is connected to x is 0.6 A, so net resistance is 10 ohms so the cell voltage is 10 × 0.6 = 6 V.

Now when M is connected to y

the net resistance is sum of the resistance of 10 Ω and 20 Ω

R_net=10+20=30 Ω

As we know by ohms law,

V=I R

Putting the values in the equation, we get

6 V= I× 30 Ω

I= = 0.2 A.

Thus, the reading of ammeter is 0.2 A.

In this circuit if we want Hot air to come out of the hair dryer then we will have to close both the switches R and S.

If we want cold air then we need to close the switch S. This will cause only the fan to work and the heater will be OFF ans thus we’ll get cold air as an output.