# Inductance in Series MCQ [Free PDF] – Objective Question Answer for Inductance in Series Quiz

1. A resistance of 7 ohms is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit.

A. 2.2A
B. 4.2A
C. 6.2A
D. 8.2A

XL = 2 × π × f × L = 10 ohm.

Impedance is given by

Z2 = (R2+XL2)

Therefore the total impedance Z = 12.2ohm.

V = IZ

∴  I = V/Z

= 100/12.2 = 8.2A.

2. A resistance of 7 ohms is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the phase difference.

A. -55.1
B. 55.1
C. 66.1
D. -66.1

The phase angle of the Inductive circuit is given as

φ = tan-1(XL/R) = 55.1

Since this is an inductive circuit, the current will lag, hence φ = -55.1.

3. A resistance of 7 ohms is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the resistor.

A. 31.8V
B. 57.4V
C. 67.3V
D. 78.2V

XL = 2 × π × f × L = 10 ohm.

Z2 = (R2+XL2)

Therefore, the total impedance Z = 12.2ohm.

V = IZ

∴I = V/Z = 100/12.2 = 8.2A.

Voltage across resistor = 8.2 × 7 = 57.4V.

4. A resistance of 7 ohms is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the inductor.

A. 52V
B. 82V
C. 65V
D. 76V

XL = 2 × π × f × L = 10 ohm. Z2 = (R2+XL2)

Therefore, the total impedance Z = 12.2ohm.

V = IZ

∴ I = V/Z = 100/12.2 = 8.2A.

Voltage across inductor = 8.2 × 10 = 82V.

5. A resistance of 7 ohms is connected in series with an inductance of 31.8mH. The circuit is connected to an x V 50Hz sinusoidal supply. The current in the circuit is 8.2A. Calculate the value of x.

A. 10V
B. 50V
C. 100V
D. 120V

XL = 2 × π × f × L = 10 ohm. Z2 = (R2+XL2)

Therefore, the total impedance Z = 12.2ohm.

V = IZ

∴ V = 12.2 × 8.2 = 100V.

6. Which, among the following, is the correct expression for φ.

A. φ = tan-1 (XL/R)
B. φ = tan-1 (R/XL)
C. φ = tan-1 (XL × R)
D. φ = cos-1 (XL/R)

From the impedance triangle, we get

tanφ = XL/R.

Hence φ = tan-1 (XL/R).

7. For an RL circuit, the phase angle is always ________

A. Positive
B. Negative
C. 0
D. 90

For a series resistance and inductance circuit, the phase angle is always a negative value because the current will always lag the voltage.

8. What is φ in terms of voltage?

A. φ = cos-1V/VR
B. φ = cos-1V × VR
C. φ = cos-1VR/V
D. φ = tan-1V/VR

From the voltage triangle, we get

cosφ = VR/V.

Hence φ = cos-1VR/V.

9. What is sinϕ from the impedance triangle?

A. XL/R
B. XL/Z
C. R/Z
D. Z/R

In the Impedance triangle, the Base is R, Hypotenuse is Z, and Height is XL.

So, sinϕ = XL/Z.

10. What is the resonance frequency of ac circuit?

A. 1/√LC
B. √(L/C.
C. √LC
D. LC