Electricity

Negative charge moves while protons can’t move because they are stuck in the nucleus so they can't move due to "strong nuclear force".

Electric current is a scalar quantity because it is the measure of how much charge flow through a particular area.

Smaller unit of current is Ampere denoted by A.

The electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field.

Given;

Charge (Q) = 5C;

Potential difference = 30 – 20 = 10V;

Work done = Charge × Potential difference

⇒ Work done = 5 × 10 = 50 J.

Hence the work done is 50 Joule.

This is because the number of the free electrons constituting current in a typical conductor is very large of order 10¹⁸ or more. This large electron number density keeps the good amount of current flowing through.

Since the wire is bend in form of the square and we know that sides of square are equal so each side will have same resistance i.e 5 ohm. Now if we make diagonal two faces in each side of the diagonal will have two sides and resistance will be 10 ohm. As shown below.

So total resistance will be

⇒ R_eq = 5 ohm

Hence equivalent resistance is 5 ohm.

Resistance do not depend on the voltage it depends on the material of the resistor whether the intermolecular force of attraction is strong or loose which decides the resistance.

In the electric fan motor the

Electrical energy is transferred into the mechanical energy, due to which some amount of the energy is lost in form of heat energy and is dissipated into the surroundings. That is why appliances like electric fan/motor become warm after continuous usage.

When the wire is cut into three equal parts the length of each part will be l/3.

And resistance is directly proportional to the length.

∴ The resistance of each part will be 9/3 = 3.

Now we have three resistors each of 3Ω connected parallel as shown in figure below.

By formula for resistors in parallel.

⇒

⇒ R_eq = 1 Ω.

Hence the equivalent resistance of the combination is 1 Ω

Given that;

Current = 0.5 A

Voltage = 220 V

Time = 30 min = 30 × 60 sec = 1800 sec

∵ 1 min = 60 sec)

We know that

⇒ Charge = Current × Time

⇒ Charge = 0.5 × 1800 C

⇒ Charge = 180 × 5 C

⇒ Charge = 900 C

Hence the charge flowing through appliance in 30 minutes is 900 Coulomb.

Given

Resistivity (ρ) = 1.7 × 10^-8 Ωm

Diameter (d) = 0.1 mm = 0.1×10^-3 m

(∵ 1m = 1000 mm)

Resistance(R) = 34 Ω

Area of cross section (A) =

A =

⇒ A =

Now

⇒

⇒

⇒

⇒

Hence the length of the wire is 15.7 m.

The schematic diagram is shown below.

Current Flowing Through 20Ω resistor

⇒

Hence the current flowing is 0.5 Ampere.

Power consumed by the 30 Ω resistor

Power = Voltage × current.

Given Voltage = 10V; current = 0.5A

⇒ Power = 10 × 0.5 = 5W

Hence power consumed by 30 Ω resistor is 5W.

(i) The above symbol represents that the two wires are not joined. As shown in the figure below.

Such symbol is used when there are two or more wires in the circuit.

(ii) The above symbol represents the Rheostat or variable resistance as Shown in figure below.

The Rheostat is used when there is need to vary the current or resistance in the circuit. The original rheostat used in the circuit is shown below in the figure below.

According to the question

Potential Difference (V) = 60V

Current drawn (I) = 4A

By Ohm’s Law

V = IR

⇒ 60 = 4 × R

⇒

Hence the resistance is 15Ω.

Given:

Resistance R = 3Ω

Original length L₁ = 15cm = 0.15m

New length, L₂ = 45 cm = 0.45 m

Formula Used:

Where, ρ is resistivity of the wire

L is the length of the wire

A is the area of the wire.

Now,

⇒

⇒

⇒ R₂ = 9 Ω

Hence, the new resistance is 9 Ω

When the wire is bent in form of a closed circle, the wires will be connected in parallel. As shown in the figure below.

Also, the resistance will become half because resistance is directly proportional to the length and the length is halved.

Resistance of each part connected in parallel = 5Ω. As shown below.

Resistance of parallel combination

⇒

⇒

Hence the resistance between ends of the diameter is 2.5Ω.

Given

Power (P) = 26W

Voltage (V) = 220V

We know that

⇒

⇒

⇒ R =

Hence the resistance is

Energy = Power × time

⇒ Energy = 26 × 10 = 260 W

We divide by 1000 for converting in Kw

Energy Consumed =

Hence the energy consumed is 0.26 KW

Both the expressions are correct. In the first case where P = I²R current (I) remains constant. Whereas in the second expression where P = V²/R the Voltage (V) remains constant.

Therefore P = I²R is used where the resistors are connected in series because in the series connection the current in each resistance is constant.

P = V²/R is used where the resistors are connected in parallel because in the parallel connection the voltage in each resistance is constant. So power becomes inversely proportional to resistance.

the voltmeter must be connected in parallel to the resistor to read the potential difference across the resistor

Voltmeter has very high resistance to ensure that it's connection do not change the flow of current in the circuit. Now if it is connected in series then no current will be there in the circuit due to its high resistance. Hence it is connected in parallel to the load across which potential difference is to be measured.

Also the voltage in the parallel connection is constant; therefore we will not get any error while measuring the voltage across resistor.

The ammeter must be connected in the series to read the current correctly. Ammeter has very low resistance If it is connected in parallel across any load then all current in circuit will choose lower resistive path (i.e ammeter) which will cause the circuit to be damaged. Hence it is used in series. Also the current in the series is always same.

Below is the circuit diagram for the voltmeter.

Below is the circuit diagram for ammeter.

Given

Resistance (R) = 2Ω

Area of cross-section = A

Length = l

We know that

⇒ (1)

New Area = A/2

New Length = 2l

New Resistance = R’

⇒

⇒

⇒

⇒ from (1).

Hence the new resistance will be four times the previous resistance i.e. 8Ω.

The resistance of the two wires are

R₁ = 2Ω and R₂ = 8Ω

When they are connected in series.

Equivalent resistance = R₁ + R₂

⇒ R_eq = 2 + 8 = 10 Ω.

R_eq in series = 10 Ω.

When they are connected in parallel.

Equivalent resistance=

⇒

⇒

⇒

R_eq in Parallel = 8/5 Ω.

Ratio of Parallel to series =

Hence the ratio of the Equivalent resistances in parallel to series is 4/25

Labelling each of the resistance as shown in the figure below.

Resistance R₁ and R₂ are in series

⇒ R_s1 = R₁ + R₂

⇒ R_s1 = 3 + 3 = 6Ω

Now R_s1 and R₃ are in parallel.

⇒

⇒

⇒

⇒

⇒ R_p = 2Ω.

Now R₄, R_p and R₅ are in series.

⇒

⇒

Hence the Equivalent resistance of the circuit is 3Ω.

For Ammeter Reading

Current (I) =

Hence the current reading in ammeter is 1A.

By Ohm’s Law

⇒ = slope of V-I graph

Therefore, the slope of theV-I graph represents the effective resistance.

Now in series combination of the resistance all the resistances get summed up

+ ………….+ R_n.

Therefore, the series combination will have the maximum resistance than the other resistors.

Hence the line with maximum slope (line C) will give the series combination of the other two resistors.

The resistors R₁ and R₂ are connected in the parallel combination because the potential difference across both the resistors is same.

Effective resistance=

⇒

⇒

⇒

Effective resistance of the combination is 2Ω.

We know that for an electric appliance

Resistance will be given as

Now, here

P = 100 W

V = 200 volts

So, R =

R =

Thus, resistance of the bulb is R = 400 Ω

Now, electric energy consumed will be

E = power of each unit × time.

⇒ E = p × t.

Here

P = 100 W

t = 4 hours

So,

E = 100 x 4 = 400 Watt per hour

Dividing by 1000 to convert into Kilo Watt per hour

E = kWh

Electrical energy consumed is kWh

The total cost of electricity = total unit of energy consumed x cost per unit =

Cost = 0.4kWh x 4 Rs/kWh = Rs1.6

Thus, total cost (daily) = Rs. 1.6

An equivalent labelled circuit for the given data is given below.

Since the voltmeter reading shows 2V. Thus V = 2.

Also, Current send by electric cell = 0.4A

Thus I = 0.4A

By Ohm’s law V = IR

⇒ 2 = 0.4 × R

⇒

The resistance of the resistor is 5Ω.

The law applicable here is Ohm’s Law.

It states that the current flowing through a conductor is directly proportional to the potential difference across the conductor

I V.

The proportionality constant is the resistance of the conductor; it is constant for the known range of the temperature.

V = IR.

The Graph obtained is the straight line with positive slope passing through the origin as shown in the figure below.

The two reasons for connecting the appliances in parallel are as follows.

1. Each appliance will be at same potential (voltage).

2. If one of the appliances fails the other will still keep working.

Given

Voltage = 220V;

Power = 1.5kW = 1500W

(1kW = 1000W)

Now

P = VI

⇒

The current in the circuit is 6.8A

Since the current in the circuit is more than the rating of the fuse (5A)the fuse will blow off.

A fuse of rating 10 A should be put off in the circuit

(i) The ammeter reading will decrease. This is because Resistance is directly proportional to the length; hence with the increase in the length the resistance of the wire will increase. So the current will decrease due to increase in the resistance (current is inversely proportional to the resistance).

(ii) The ammeter reading will increase because.

Thus the resistance is inversely proportional to the area of the cross section, so with the increase in the area the resistance of the wire will decrease.

Therefore with the decrease in the resistance the current will increase (current is inversely proportional to the resistance).hence the ammeter reading will increase

Given that the potential difference is 1 V. The potential difference is the Difference in the potential between two points that represents the work involved or energy released in transfer of a unit quantity of electricity from one point to the other. The figure below illustrates it

V_A = potential at A

V_B = potential at B

From the above definition of the potential difference the above statement means that 1Joule of work is being done to move a charge of 1 coulomb from point A to point B.

Resistors across AC and DE are in parallel i.e. R_AC and R_DE are in parallel.

So

⇒

⇒

⇒

Now R’_p and R_BC are I series.

R’_s = R’_p + R_BC

⇒ R’_s = 15 + 15 = 30Ω.

Again, R_AB and R’_s are in parallel.

⇒

⇒ R’’_p =

Equivalent resistance of the above circuit is 10Ω.

Now, current flowing through the circuit is

I =

The value of current is 0.3A

Two parallel combinations must be connected in series with each other to get the effective resistance of 8Ω. As shown in the figure below.

The effective resistance of each of the parallel combination is 4Ω. And this two 4Ω resistors are added together to get 8 Ω effective resistance.

Shushant told his teacher that ammeter was broke by him he did not hide from his teacher this shows that Shushant is very honest and respectful towards his teacher. Also, in spite of his group mates telling him not to tell to the teacher he developed enough courage to tell everything to his teacher this show that he is courageous and confident student who accept his mistake.

An ammeter is used to measure

the electric current in the unit of Ampere (A).It can also be used to verify the Ohm’s law.

It is always connected in the series. The figure below shows the ammeter and its connection in the circuit.

Ammeter

Ammeter connected in series.

The ammeter is connected in the series because it has very low resistance If it is connected in parallel across any load then all current in circuit will choose lower resistive path (i.e. ammeter) which will cause the circuit to be damaged. Hence it is used in series. Also the current in the series is always same.

Shushant could have used the ammeter to verify the ohm’s law which states that “current flowing through a conductor is directly proportional to the potential difference across the conductor”

I V.

After setting up the circuit (as shown in figure below) and drawing the graph between V and I will get the straight line passing through origin.

The Ammeter will measure the current and the Voltmeter will measure the potential difference across the resistor and the ratio of voltage and current will give the resistance which is constant.