Electric Current Practice Questions

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Important Questions on Electric Current

Ques 1. Define the mobility of charge carriers in a conductor. Write its SI unit also. (2 Marks)

Ans. The magnitude of the drift velocity of charge carriers per unit applied electric field is called mobility of charge carriers. Its SI unit is m²V^-1s^-1. 

Ques 2. A 10 V cell of negligible Internal resistance Is connected In parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure. Find the value of current in the circuit. (2 Marks)

Ans. Given E₁ = 10 V, r₁ = 0, E₂ = 200 V, r₂ = 38 Ω

The two cells send current in the opposite direction, therefore, net emf of the combination

E_net = 200 – 10 = 190 V

Hence, current in the circuit is

I = E_net / R 

I = 190 / 38 

I = 5 A

Ques 3. The circuit in the figure shows two cells connected in opposition to each other. Cell E₁ is of emf 6 V and internal resistance 2 Ω; the cell E₂ is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B. (3 Marks)

Ans. The cells in the circuit shown above are connected in opposite directions, therefore the net EMF is;

E = E₁ - E₂ 

E = 6 - 4 = 2 V

Hence the current in the circuit is,

I = E / R + r

I = 2 / 10 

I = 0.2 A

Potential difference across E₁ = 6 - 0.2 × 2 = 5.6 V

Potential difference across E₂ = VAB = 4 + 0.2 × 8 = 5.6 V

Ques 4. If an electric field of magnitude 570 NC^-1 is applied in the copper wire, find the acceleration experienced by the electron. (3 Marks)

Ans. E = 570 NC^-1, e = 1.6 × 10^-19 C, m = 9.11 × 10^-31 kg, a = ?

F = ma = eE

a = eE / m 

a = 570 × 1.6 × 10^-19 / 9.11 × 10^-31

a = 912 × 10^-19 × 10³¹ / 9.11 

a = 1.001 × 10¹⁴ ms^-2 

Ques 5. A copper wire of cross-sectional area 0.5 mm² carries a current of 0.2 A. If the free electron density of copper is 8.4 × 10²⁸ m^-3 then compute the drift velocity of free electrons. (3 Marks)

Ans. The relation between the drift velocity and current in the wire is given by;

V_d = I / ne A

V_d = 0.2 / 8.4 × 10²⁸ × 1.6 × 10^-19 × 0.5 × 10^-6

V_d = 0.03 × 10^-3 ms^-1 

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Ques 6. Determine the number of electrons flowing per second through a conductor when a current of 32 A flows through it. (3 Marks)

Ans. I = 32 A, t = 1 sec

Charge of an electron, e = 1.6 × 10^-19 C

Let n be the number of electrons flowing per second

I = Q / t

I = ne / t

n = It / e

n = 32 × 1 / 1.6 × 10^-19

n = 20 × 10¹⁹ 

n = 2 × 10²⁰ electrons

Ques 7. Calculate the equivalent resistance for the circuit which is connected to 24 V battery and also find the potential difference across 4 Ω and 6 Ω resistors connected in series in the circuit. (3 Marks)

Ans. The two resistors are connected in series, therefore the effective resistance in the circuit is

Req = 4 Ω + 6 Ω = 10 Ω

Current in the circuit is, I = V / R_eq = 24 / 10 = 2.4 A

Voltage across 4 Ω resistor,

V₁ = IR₁ = 2.4 × 4 = 9.6 V

V₂ = IR₂ = 2.4 × 6 = 14.4 V

Ques 8. Calculate the equivalent resistance in the following circuit and also find the current I, I₁ and I₂ in the given circuit. (5 Marks)

Ans. The resistances are connected in parallel, so the net resistance is given by; 

1/R_p = 1/R₁ + 1/R₂ 

1/R_p = 1/4 + 1/6 

1/R_p = 5/12 Ω

R_p = 12/5 Ω

The resistors are connected in parallel and the voltage across each resistor is same, so

I₁ = V/R₁ = 24/4 = 6 A

I₂ = V/R₂ = 24/6 = 4 A

Total Current, I = I₁ + I₂ = 6 + 4 = 10 A

Ques 9. Find the value of I from the given circuit. (3 Marks)

Ans. Applying Kirchhoff’s rule to the center point of the circuit. The arrows pointing towards the center are positive and those away from the center are negative. 

So, 0.2 A - 0.4 A + 0.6 A - 0.5 A + 0.7 A - I = 0 

1.5 A - 0.9 A - I = 0

0.6 A - I = 0

I = 0.6 A

Ques 10. Calculate the current that flows in the 1 Ω resistor in the following circuit. (5 Marks)

Ans. 

From the above figure, we can say that the current flowing through the 9 V battery as I₁ splits into I₂ and I₁ - I₂ (according to Kirchhoff’s current rule).

Consider the loop EFCBE, and apply KVR, we get

1I₂ + 3I₁ + 2I₁ = 9

5I₁ + I₂ = 9 ……..(1)

Applying KVR to the loop EADFE, we get

3 (I₁ - I₂) - 1I₂ = 6

3I₁ - 4I₂ = 6 ………(2)

Solving eq. (1) and (2), we get

I₁ = 1.83 A

I₂ = -0.13 A

The negative sign indicates that the current in the 1Ω resistor flows from F to E. 

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