# Physics Questions and Answers – AC Voltage Applied to an Inductor

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “AC Voltage Applied to an Inductor”.

1. What is the reactance of an inductor in a dc circuit?
a) Maximum
b) Minimum
c) Zero
d) Indefinite

Explanation: The reactance of an inductor in a dc circuit is zero.
For dc ➔ f = 0
XL = 2π f L
XL = 0
Therefore, the reactance of an inductor in a dc circuit is zero.

2. In which type of circuit the value of power factor will be minimum?
a) Resistive
b) Inductive
c) Superconductive
d) Semi conductive

Explanation: For a purely inductive circuit, Φ = ± $$\frac {\pi }{2}$$
Power factor = cos (±$$\frac {\pi }{2}$$)
Power factor = 0
Therefore, the power factor will be minimum for an inductive circuit.

3. The frequency of ac is doubled. How does XL get affected?
a) XL gets doubled
b) XL becomes zero
c) XL is halved
d) XL is indefinite

Explanation: When the frequency of an ac is doubled ➔ The inductive reactance (XL) gets doubled.
This is because inductive reactance is directly proportional to the frequency of an alternating current circuit.

4. At what frequency will a coil, which has an inductance of 2.5 H, have a reactance of 3500 Ω?
a) 700Hz
b) 350 Hz
c) 200 Hz
d) 223 Hz

Explanation: f = $$\frac {X_L}{2\pi L}$$
f = $$\frac {3500}{(2 \times 3.14 \times 2.5)}$$
f = 222.9 Hz ≈ 223 Hz
Therefore, the frequency of a coil having reactance of 3500 Ω is 223 Hz.

5. Why does an inductor offer an easy path to dc and a resistive path to ac?
a) XL is maximum for dc and infinite for ac
b) XL is zero for dc and infinite for ac
c) XL is zero for dc and finite for ac
d) XL is maximum for dc and finite for ac

Explanation: For dc, f = 0,
XL = 2π f L
XL = 0
XL is zero for dc and has a finite value for ac Hence an inductor offers an easy path to dc and a resistive path to ac.

6. The inductive reactance for an ac circuit containing only an inductor is $$\frac {E_0}{X_L}$$.
a) True
b) False

Explanation: No, this statement is false. The inductive reactance for an ac circuit containing only an inductor is XL = ωL.
XL = ωL
XL = 2π f L

7. Determine the peak current if an inductor of inductance 500 mH is connected to an ac source of peak emf 650 V and frequency 100 Hz
a) 1.55 A
b) 2.07 A
c) 7.89 A
d) 9.87 A

Explanation: Peak current (I0) = $$\frac {E_0}{X_L}$$
I0 = $$\frac {650}{(2 \times 3.14 \times 100 \times 0.5)}$$
I0 = 2.07 A
Therefore, the peak current is calculated as 2.07 A.

8. Find out the rms value of current in the circuit wherein a 35 mH inductor is connected to 200 V, 70 Hz ac supply.
a) 13 A
b) 15 A
c) 20 A
d) 45 A

Explanation: XL = 2π f L
XL = 2π × 70 × 35 × 10-3 Ω.
Irms = $$\frac {E_{rms}}{X_L} = \frac {200}{(2\pi \times 70 \times 35 \times 10^{-3})}$$
Irms = 12.99 A ≈ 13 A
Therefore, the rms value of current in the circuit is 13 A.

Sanfoundry Global Education & Learning Series – Physics – Class 12.