Current Electricity MCQs with Solutions

Ans. (c). 1 coulomb of charge flowing per second

Explanation: Electric current is the rate of flow of charge in a conductor. 

I = Q / t

where; 

I = current

Q = charge (in coulombs)

t = time (in sec)

If the value of Q is 1 coulomb and time is 1 second, then

I = 1 C / 1 sec

I = 1 Cs^-1 

1 A = 1 Cs^-1 

Ans. (d). 600 Coulomb

Explanation: We know, I = Q / t

Here, I = 2 A and t = 5 minutes = 300 sec

So, Q = I × t 

Q = 2 × 300 

Q = 600 C

Ans. (b). 6.25 × 10¹⁵

Explanation: I = Q / t 

Given, I = 1 mA = 1 × 10^-3 A and t = 1 sec

Total charge passing through the conductor per second is given by,

Q = ne

ne = Q

ne = I × t

n = I × t / e

n = 1 × 10^-3 × 1 / 1.6 × 10^-19 

n = 6.25 × 10¹⁵ 

Ans. (d). 0.8 mA

Explanation: Given, r (radius) = 0.5 × 10^-10 m and f (frequency) = 5 × 10¹⁵ s^-1 

I = Q / t 

And we know, frequency (f) = 1 / t

∴ I = Q × f

I = 1.6 × 10^-19 × 5 × 10¹⁵ = 8 × 10^-4 A = 0.8 mA

Ans. (c). direction of conventional current in metallic conductors

Explanation: The direction of conventional current is always in the direction of flow of positive charges (from higher potential to lower potential). Electric current is due to flow of electrons i.e. from lower to higher potential. Hence, the direction of electric current is always opposite to the direction of conventional current in metallic conductors. 

Ans. (b). 750 C

Explanation: Given t = 5 minutes = 5 × 60 sec

Voltage of the cell (V) = 5 volt

Resistor (R) = 2Ω

V = IR

I = V / R = 5 / 2 = 2.5 A

Q (Electric Charge) = I × t = 2.5 × 5 × 60 = 750 C

Ans. (a). 2.20Ω

Explanation: Given, R₁ = 9Ω, R₂ = 7Ω and R₃ = 5Ω

In parallel combination, the net resistance is given by; 

1 / R_eq = 1/9 + 1/7 + 1/5 

1 / R_eq = 35 + 45 + 63/315 

1 / R_eq = 143/315

R_eq = 315/143 = 2.20Ω

Ans. (a). 10²⁰ 

Explanation: Given I = 1 A ; t = 16 sec ; e = 1.6 × 10^-19 C

I = Q / t 

I = Ne / t

N = I × t / e (where, N = no. of electrons)

N = 1 A × 16 sec / 1.6 × 10^-19 C = 10²⁰ 

Ans. (a). 24 C

Explanation: Instantaneous current across a conductor is defined as; 

I = dQ / dt

dQ = I × dt

dQ = (2t + 3t²) dt …….(i)

Now, integrate eq. (i)

∫ dQ = ∫ (2t + 3t²) dt

ΔQ = 2 [t²/2] + 3 [t³/3]

ΔQ = [9 - 4] + [27 - 8] 

ΔQ = 24 C

Ans. (b). 6.25 × 10¹⁹

Explanation: Q = It = 5 × 2 = 10 C

Charge on one electron = 1.6 × 10^-19  C

N = no. of electrons

Q = Ne 

N = Q/e

N = 10 / 1.6 × 10^-19 

N = 6.25 × 10¹⁹ 

Ans. (b). Ammeter

Explanation: Ammeter is the device used to measure the electric current flowing through a circuit. The resistance of the ammeter is very low, ideally zero, thus it allows all the current to pass through it and measures it. 

Voltmeter, on the other hand, is an instrument used to measure voltage across two terminals in a circuit. A Galvanometer is used to detect the direction and magnitude of small current present in the circuit. It is used when magnetic fields are present and measures only direct current. 

Potentiometer is used to determine the emf and internal resistance of a cell. It measures the unknown voltage by comparing it with the known voltage.