Vidaara.orgClass 12 · Physics
CodeVID-P12-03-INA-01
Inductance & AC Circuits — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all steps in numericals and draw neat phasor diagrams wherever asked. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Inductive reactance $X_L$ equals:
- A.$1/\omega L$
- B.$\omega L$
- C.$\omega/L$
- D.$L/\omega$
2.
Capacitive reactance is largest at:
- A.high frequency
- B.low frequency
- C.resonance
- D.any frequency
3.
$I_{rms}$ equals:
- A.$I_0$
- B.$I_0/\sqrt2$
- C.$I_0\sqrt2$
- D.$2I_0$
4.
Power factor of a pure inductor is:
- A.1
- B.0.5
- C.0
- D.0.707
5.
At resonance the LCR current is:
- A.minimum
- B.maximum
- C.zero
- D.unchanged
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define self-inductance and give its unit.
7.
A 0.1 H inductor is in a 50 Hz circuit. Find $X_L$.
8.
What is wattless current?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Define impedance and write its expression for a series LCR circuit.
10.
A series LCR circuit has $R = 8\,\Omega$, $X_L = 16\,\Omega$, $X_C = 10\,\Omega$. Find Z.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Explain series resonance in an LCR circuit using a phasor diagram, derive $\omega_0 = 1/\sqrt{LC}$, and find $f_0$ for $L = 0.5\,\text{H}$, $C = 2\,\mu\text{F}$.
Answer Key
Section A — Multiple Choice Questions
- (B) $\omega L$
- (B) low frequency
- (B) $I_0/\sqrt2$
- (C) 0
- (B) maximum
Section B — Short Answer (2 marks)
- $\varepsilon = -L\,dI/dt$; opposition to change of current; unit henry (H).
- $X_L = \omega L = 2\pi \times 50 \times 0.1 \approx 31.4\,\Omega$.
- Current in a pure L or C that consumes no average power since $\cos\phi = 0$.
Section C — Short Answer (3 marks)
- Net AC opposition; $Z = \sqrt{R^2 + (X_L - X_C)^2}$, with $I_{rms} = V_{rms}/Z$.
- $Z = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\,\Omega$.
Section D — Long Answer (5 marks)
- At resonance $X_L = X_C$, $Z = R$ minimum; $\omega_0 = 1/\sqrt{LC} = 1000\,\text{rad/s}$; $f_0 = \omega_0/2\pi \approx 159\,\text{Hz}$.
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