250+ TOP MCQs on AC Voltage Applied to a Series LCR Circuit | Class12 Physics

Physics Exam Questions on “AC Voltage Applied to a Series LCR Circuit”.

1. Determine the impedance of a series LCR-circuit if the reactance of C and L are 250 Ω and 220 Ω respectively and R is 40 Ω.
a) 250 Ω
b) 150 Ω
c) 50 Ω
d) 80 Ω
Answer: c
Clarification: Z = √(R² + (X_L – X_C)²
Z = √40² + (220 – 250)²
Z = 50 Ω.s

2. If a 0.5 H inductor, 80 μF capacitor and a 40 Ω resistor are connected in series with a 150 V, 60 Hz supply. Calculate the impedance of the circuit.
a) 100 Ω
b) 160.3 Ω
c) 50 Ω
d) 65 Ω
Answer: b
Clarification: Impedance, Z = √(R² + (X_L – X_C)².
Z = √40² + (188.4 – 33.17)²
Z = 160.3 Ω

3. An inductor coil joined to a 6 V battery draws a steady current of 12 A. This coil is connected in series to a capacitor and ac source of alternating emf 6 V. If the current in the circuit is in phase with the emf, find the rms current.
a) 12 A
b) 15 A
c) 25 A
d) 19 A
Answer: a
Clarification: R = (frac {V}{I} = frac {6}{12}) = 0.5 Ω.
Impedance = Z = R = 0.5 Ω,
I_rms = (frac {E_{rms}}{Z})
I_rms = (frac {6}{0.5})
I_rms = 12 A.

4. A resistor of 50 Ω, an inductor of (frac {20}{pi}) H and a capacitor of (frac {5}{pi }) μF are connected in series to a voltage supply of 230 V – 50 Hz. Find the impedance of the circuit.
a) 150 Ω
b) 250 Ω
c) 350 Ω
d) 50 Ω
Answer: d
Clarification: X_L = 2πfL
X_L = 2π × 50 × ((frac {20}{pi })) = 2000 Ω.
X_C = (frac {1}{2πfC})
X_C = 2000 Ω.
Z = √(R² + (X_L – X_C)²
Z = √50² + (2000 – 2000)²
Z = 50 Ω.

5. A series circuit with L = 0.12 H, C = 0.48 mF and R = 25 Ω is connected to a 220 V variable frequency power supply. At what frequency is the circuit current maximum?
a) 79 Hz
b) 19 Hz
c) 21 Hz
d) 93 Hz
Answer: c
Clarification: f_r = (frac {1}{( 2pi sqrt {LC} )})
f_r = (frac {1}{( 2 times 3.14 times sqrt {0.12} times 0.48 times 10^{-3} )})
f_r = 21Hz.

6. Voltages across L and C in series are 90° out of phase.
a) True
b) False
Answer: b
Clarification: Given a current in series LC, the voltage in inductor (L) leads current by 90° phase and voltage in capacitor (C) lags behind current by 90° phase. So voltage in inductor and capacitor differ by a phase of 180°.

7. A capacitor, resistor of 10 Ω, and an inductor of 70 mH are in series with an ac source marked 120 V, 70 Hz. If it is found that voltage is in phase with the current, then find out the impedance of the circuit.
a) 5 Ω
b) 10 Ω
c) 100 Ω
d) 900 Ω
Answer: b
Clarification: X_C = X_L.
C = (frac {1}{4π^2f^2L})
C = (frac {1 times 7 times 7}{(4 times 22 times 22 times 70^2 times 70 times 10^{-3} )})
C = 7.37 × 10^-5 F.
Impedance (Z) = √(R² + (X_L – X_C)²
Z = 10 Ω.

8. When are the voltage and current in LCR-series ac circuit in phase?
a) X_L = X_C
b) X_L > X_C
c) X_L < X_C
d) Indeterminant
Answer: a
Clarification: When X_L = X_C, the resultant vector for net reactance will be zero and the value of θ will also be zero. Hence, when current and voltage will be in the same phase, it is called as a purely resistive circuit and in this case the peak current will be maximum.

9. What is the reactance of an inductor in a dc circuit?
a) Minimum
b) Maximum
c) Zero
d) Indefinite
Answer: c
Clarification: For dc ➔ f = 0
So X_L = 2πfL
X_L = 2 × π × 0 × L
X_L = 0
Therefore, the reactance of an inductor in a dc circuit is zero.

10. If the frequency of the ac source in a series LCR-circuit is increased, how does the current in the circuit change?
a) Decreases then increase
b) Increases then decrease
c) Becomes zero
d) Remains constant
Answer: b
Clarification: With the increase in frequency, the current in a series LCR circuit undergoes a series of changes, i.e. the current in a series LCR-circuit first increases, attains a maximum value (at f = f_r) and then decreases.

Physics Exam Questions and Answers for Class 12,