Unit of Voltage Important Questions

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The unit of voltage in the SI system is Volt (V). Voltage is the potential difference present between the two points of the circuit. It is also defined as the work done per unit charge to move it from one point to the other in an electric circuit. The dimensional formula of voltage is [ML²T^-3I^-1]. The formula of voltage is;

V = IR

The voltage across any two points of a circuit is measured with the help of an instrument called Voltmeter. To measure the potential difference with a voltmeter, one electrical lead of the voltmeter is connected to the first point and other to the second point. When there is a difference between the measurements at each terminal of the device with respect to the reference point, it is called voltage drop.

A Voltmeter

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Important Questions on Unit of Voltage

Ques 1. The figure shows the V – l graph for a parallel and series combination of two resistors A and B. Which line represents the parallel combination? (2 Marks)

The figure shows the V – l graph for a parallel and series combination of two resistors A and B

Ans. R = V/I

Slope of B > Slope of A

For the same potential, the current is less in series combination than parallel combination. Therefore from the graph, it is apparent that the same potential current is less in A. Therefore, B represents the parallel combination.

Ques 2. Two cells of emf 4 V and 2 V and internal resistance 2 Ω and 1 Ω respectively are connected in parallel so as to send the current in the same direction through an external resistance of 10 Ω. Find the potential difference across 10 Ω resistor. (3 Marks)

Ans. Equivalent internal resistance in parallel connection,

r_eq= r₁r₂ / r₁ + r₂

r_eq= 1 x 2 / 1 + 2

r_eq= ⅔ Ω

Equivalent emf of cells in parallel combination,

E_eq= [ E₁/r₁ + E₂/r₂ ] r_eq

E_eq= [ 2/1 + 4/2 ] ⅔ = 8/3 V

Potential difference across 10 Ω resistor,

E = (R / R + r_eq) E_eq

∴ E = (10 / 10 + ⅔ ) 8/3 = 2.5 V

Ques 3. A battery of emf 10 V and internal resistance 3Ω is connected to a resistor. The current in the circuit is 0.5 A. What is the terminal voltage of the battery when the circuit is closed? (3 Marks)

Ans. Emf of battery (E) = 10 V, internal resistance of the battery (r) = 3 , current (I) = 0.5 A

Let resistance of the resistor be R

I = E / (R + r)

R + r = E/I

R + r = 10/0.5

R = 20 - 3 = 17Ω

Terminal Voltage of the resistor = V

V = IR

V = 0.5 17

V = 8.5 V

The terminal voltage of the battery is 8.5 V.

Ques 4. A storage battery of emf 8.0 V and internal resistance 0.5Ω is being charged by a 120 V DC supply using a series resistor of 15.5Ω . What is the terminal voltage of the battery during charging? (3 Marks)

Ans. E = 8 V, internal resistance (R₁) = 0.5Ω

V = 120 V, resistance (R₂) = 15.5Ω

Effective voltage (V’) = V - E = 112 V

I = V’ / (R₁ + R₂)

I = 7 A

Voltage across R₁ = IR₁ = 3.5 V

Terminal voltage = 8 + 3.5 = 11.5 V

Ques 5. If the discharged battery is charged by an external emf source, is the terminal voltage of the battery during charging greater or less than its emf 12 V? (3 Marks)

Ans. During charging, the current flows in the opposite direction i.e. from positive to negative terminal inside the cell. So, during charging,

V = E + IR

V > E = 12 V

Hence, the terminal voltage of the battery during charging is greater than its emf.

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Ques 6. Two batteries of emf 4 V and 8 V with internal resistance 1Ω and 2Ω respectively are connected to an external resistance R = 9Ω as shown in figure. What is the current and the potential difference between P and Q respectively? (3 Marks)

Two batteries of emf 4 V and 8 V with internal resistance 1Ω and 2Ω respectively are connected to an external resistance R = 9Ω as shown in figure

Ans. In the figure given above, the batteries are connected in reverse polarities, the net potential applied to the circuit = 8V - 4V = 4V

The net resistance in the circuit = R + r₁ + r₂ = 9 + 1 + 2 = 12Ω

Current in the circuit = 4/12 = ⅓ A

Potential difference across P and Q = IR = ⅓ 9 = 3 V

Ques 7. Two batteries, one of emf 18 V and internal resistance 2 and the other of emf 12 V and internal resistance 1 , are connected as shown. What will be the reading in the voltmeter V? (3 Marks)

Two batteries, one of emf 18 V and internal resistance 2 and the other of emf 12 V and internal resistance 1 , are connected as shown

Ans. Here the batteries are connected in parallel,

V = E₁r₂ + E₂r₁ / r₁ + r₂

V = (18 x 1 + 12 x 2) / 1 + 2

V = 14 V

Ques 8. A cell having an emf ϵ and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, draw the plot of potential difference V across R. (5 Marks)

Ans. The circuit diagram is as follows:

The circuit diagram

Now, applying KVL in the loop,

ϵ - iR - ir = 0

i = ϵ / R + r

V_{across R} = iR

V = ϵR / R + r

V = ϵ / (1 + r/R)

So, at R = 0, V = 0

and when R = ∞, then V = ϵ

So, the graph of V versus R will be:

Ques 9. A battery of emf 6 V and internal resistance 2Ω is connected to a resistor. If the current in the circuit is 0.25 A, find the terminal voltage of the battery. (5 Marks)

Ans. Given I = 0.25 A, emf (ε) = 6 V, internal resistance (r) = 2

I = ε / (r + R)

0.25 = 6 / (2 + R)

2 + R = 600/25

2 + R = 24

Resistance, R = 24 - 2 = 22

Now, V = IR

V = 25 22 / 100

V = 22/4

V = 5.5 V

Hence, the terminal voltage is 5.5 V

Ques 10. The figure shows a plot of terminal voltage ‘V’ versus the current ‘i’ of a given cell. Calculate from the graph

(a) emf of the cell and

(b) internal resistance of the cell. (5 Marks)

internal resistance of the cell

Ans. From the graph,

(a). E = V = 6 V (when I = 0)

(b). E = V + Ir

From the graph, when I = 1 A, V = 4 V

6 = 4 + 1 (r)

r = 2 Ω

Ques 11. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E? (5 Marks)

Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A

Ans. The resistors 5Ω and 10Ω are in series,

∴ R_s = 5 + 10 = 15Ω

Now, 15 and 10 are in parallel,

So, R_P’ = (15 10) / 25 = 6Ω

R_P’ and 30 are also in parallel,

So, R_P’’ = (6 30) / 36 = 5Ω

But R_p’’, R and 15Ω are in series,

R_P’ = R_P’’ + R + 15 Ω

R_P’ = 5 + R + 15

R_P’ = 20 + R

Current = 0.2 A

ε / Total resistance = Total current

(8 - 3) / (20 + R) = 0.2

R = 5Ω

Potential difference between points B and E

= I R_BE = 0.2 x 5 = 1 V

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