December 11, 2025
10 Most Common Alternating Current Problems with Step-by-Step Solutions
Numerical Problems

10 Most Common Alternating Current Problems with Step-by-Step Solutions (Class 12, JEE Advanced & NEET)

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Master high-difficulty Alternating Current problems with key formulas, RMS & phasor analysis, resonance, and power calculations. Step-by-step solutions for JEE, NEET, and CBSE Class 12.

Alternating Current (AC) isn’t just about plugging numbers into formulas—it’s about understanding phase, impedance, energy flow, and resonance in dynamic circuits. While NCERT covers basics, JEE Advanced and NEET often test layered applications involving combinations of R, L, and C under non-ideal or time-varying conditions.

This post gives you 10 carefully selected, high-difficulty AC problems—each designed to reflect real competitive exam patterns—with crystal-clear, step-by-step reasoning. Plus, we’ve included a quick-reference formula sheet at the top so you can revise and apply concepts instantly.

Contents hide
1 ⚡ Essential Alternating Current Formulas (Class 12 & JEE)
2 Problem 1: RMS & Instantaneous Current with Phase Shift
3 Problem 2: Impedance Triangle & Unknown Component
4 Problem 3: Power Factor Correction
5 Problem 4: Series LCR with Variable Frequency
6 Problem 5: Quality Factor & Bandwidth
7 Problem 6: Parallel LCR Circuit Analysis
8 Problem 7: AC Source with Internal Resistance
9 Problem 8: Transformer with Load & Efficiency
10 Problem 9: Time-Dependent Power & Energy
11 Problem 10: Complex Impedance & Phasor Diagram
12 Final Advice for Competitive Exams
13 🔗Go Deeper on PhysicsQandA.com

⚡ Essential Alternating Current Formulas (Class 12 & JEE)

Keep this handy while solving:

  • Peak ↔ RMS: V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}, \quad I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}}
  • Angular frequency: \omega = 2\pi f
  • Inductive reactance: X_L = \omega L
  • Capacitive reactance: X_C = \dfrac{1}{\omega C}
  • Impedance (series): Z = \sqrt{R^2 + (X_L - X_C)^2}
  • Phase angle: \tan \phi = \dfrac{X_L - X_C}{R}
  • Power factor: \cos \phi = \dfrac{R}{Z}
  • Average power: P = V_{\text{rms}} I_{\text{rms}} \cos \phi = I_{\text{rms}}^2 R
  • Resonant frequency: f_0 = \dfrac{1}{2\pi\sqrt{LC}}
  • Quality factor (Q): Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 C R}

Problem 1: RMS & Instantaneous Current with Phase Shift

Problem: The instantaneous voltage of an AC source is v = 311 \sin(314t) volts. It’s connected to a circuit where current is i = 2.0 \sin(314t - \frac{\pi}{6}) A. Find (a) RMS voltage and current, (b) frequency, and (c) average power consumed.

RMS & Instantaneous Current with Phase Shift

Solution:
(a) V_{\text{rms}} = \dfrac{311}{\sqrt{2}} \approx 220\, \text{V}, \quad I_{\text{rms}} = \dfrac{2.0}{\sqrt{2}} \approx 1.414\, \text{A}
(b) \omega = 314 = 2\pi f \Rightarrow f = \dfrac{314}{2\pi} \approx 50\, \text{Hz}
(c) Phase difference \phi = \frac{\pi}{6} = 30^\circ, so \cos\phi = \frac{\sqrt{3}}{2} \approx 0.866
Average power P = V_{\text{rms}} I_{\text{rms}} \cos\phi = 220 \times 1.414 \times 0.866 \approx 270\, \text{W}

Problem 2: Impedance Triangle & Unknown Component

Problem: A series circuit draws 1.0 A RMS from a 100 V, 50 Hz source. The power dissipated is 80 W, and the circuit is inductive. Find (a) resistance, (b) reactance, and (c) inductance.

Impedance Triangle

Solution:
(a) P = I^2 R \Rightarrow 80 = (1)^2 R \Rightarrow R = 80\, \Omega
(b) Z = \dfrac{V}{I} = \dfrac{100}{1} = 100\, \Omega
Since Z = \sqrt{R^2 + X_L^2} \Rightarrow 100 = \sqrt{80^2 + X_L^2}
\Rightarrow X_L = \sqrt{100^2 - 80^2} = \sqrt{3600} = 60\, \Omega
(c) X_L = 2\pi f L \Rightarrow L = \dfrac{60}{2\pi \times 50} \approx \dfrac{60}{314} \approx 0.191\, \text{H}

Problem 3: Power Factor Correction

Problem: A 220 V, 50 Hz motor draws 10 A with a power factor of 0.6 lagging. What capacitance must be added in parallel to raise the power factor to 0.9?

Power factor correction

Solution:
Initial apparent power: S_1 = V I = 220 \times 10 = 2200\, \text{VA}
Real power (constant): P = S_1 \cos\phi_1 = 2200 \times 0.6 = 1320\, \text{W}
New apparent power: S_2 = \dfrac{P}{\cos\phi_2} = \dfrac{1320}{0.9} \approx 1466.7\, \text{VA}
Initial reactive power: Q_1 = \sqrt{S_1^2 - P^2} = \sqrt{2200^2 - 1320^2} = 1760\, \text{VAR}
New reactive power: Q_2 = \sqrt{S_2^2 - P^2} \approx \sqrt{1466.7^2 - 1320^2} \approx 641\, \text{VAR}
Capacitor must supply: Q_C = Q_1 - Q_2 = 1760 - 641 = 1119\, \text{VAR}
But Q_C = V^2 \omega C \Rightarrow C = \dfrac{Q_C}{V^2 \cdot 2\pi f} = \dfrac{1119}{(220)^2 \cdot 314} \approx \dfrac{1119}{15,185,600} \approx 73.7\, \mu\text{F}

Problem 4: Series LCR with Variable Frequency

Problem: In a series LCR circuit, R = 20\, \Omega, L = 100\, \text{mH}, C = 50\, \mu\text{F}. The AC source voltage is 100 V (RMS). Find (a) resonant frequency, (b) current at resonance, and (c) current when frequency is 1.5× resonance.

Resonance

Solution:
(a) f_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{0.1 \times 50 \times 10^{-6}}} = \dfrac{1}{2\pi\sqrt{5 \times 10^{-6}}} \approx \dfrac{1}{0.014} \approx 71.2\, \text{Hz}
(b) At resonance, Z = R = 20\, \Omega \Rightarrow I_0 = \dfrac{100}{20} = 5\, \text{A}
(c) New f = 1.5 \times 71.2 \approx 106.8\, \text{Hz}
X_L = 2\pi f L = 2\pi \times 106.8 \times 0.1 \approx 67.1\, \Omega
X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \times 106.8 \times 50 \times 10^{-6}} \approx 29.8\, \Omega
Z = \sqrt{20^2 + (67.1 - 29.8)^2} = \sqrt{400 + (37.3)^2} = \sqrt{400 + 1391} \approx \sqrt{1791} \approx 42.3\, \Omega
I = \dfrac{100}{42.3} \approx 2.36\, \text{A}

Problem 5: Quality Factor & Bandwidth

Problem: A series LCR circuit has L = 2\, \text{H}, C = 32\, \mu\text{F}, R = 10\, \Omega. Calculate (a) resonant frequency, (b) Q-factor, and (c) bandwidth.

Quality factor bandwidth

Solution:
(a) f_0 = \dfrac{1}{2\pi\sqrt{2 \times 32 \times 10^{-6}}} = \dfrac{1}{2\pi\sqrt{64 \times 10^{-6}}} = \dfrac{1}{2\pi \times 8 \times 10^{-3}} \approx 19.9\, \text{Hz}
(b) Q = \dfrac{1}{R} \sqrt{\dfrac{L}{C}} = \dfrac{1}{10} \sqrt{\dfrac{2}{32 \times 10^{-6}}} = 0.1 \times \sqrt{62,500} = 0.1 \times 250 = 25
(c) Bandwidth \Delta f = \dfrac{f_0}{Q} = \dfrac{19.9}{25} \approx 0.796\, \text{Hz}

💡 High Q = sharp resonance—key for tuning circuits in radios!

Problem 6: Parallel LCR Circuit Analysis

Problem: A 100 Ω resistor, 0.2 H inductor, and 10 µF capacitor are connected in parallel to a 100 V, 50 Hz source. Find total RMS current and phase angle.

Parallel LCR circuit

Solution:
Branch currents:
– Resistor: I_R = \dfrac{100}{100} = 1.0\, \text{A} (in phase)
– Inductor: X_L = 2\pi \times 50 \times 0.2 = 62.8\, \Omega \Rightarrow I_L = \dfrac{100}{62.8} \approx 1.59\, \text{A} (lags by 90°)
– Capacitor: X_C = \dfrac{1}{2\pi \times 50 \times 10^{-5}} \approx 318.3\, \Omega \Rightarrow I_C = \dfrac{100}{318.3} \approx 0.314\, \text{A} (leads by 90°)
Net reactive current: I_X = I_L - I_C = 1.59 - 0.314 = 1.276\, \text{A} (inductive)
Total current: I = \sqrt{I_R^2 + I_X^2} = \sqrt{1^2 + 1.276^2} \approx \sqrt{2.628} \approx 1.62\, \text{A}
Phase angle: \tan \phi = \dfrac{I_X}{I_R} = 1.276 \Rightarrow \phi \approx 52^\circ (lagging)

Problem 7: AC Source with Internal Resistance

Problem: An AC generator has EMF \varepsilon = 200 \sin(100\pi t) V and internal resistance r = 5\, \Omega. It powers a load of R = 15\, \Omega, L = 0.1\, \text{H} in series. Find power delivered to the load.

AC Generator

Solution:
\omega = 100\pi \Rightarrow f = 50\, \text{Hz}
Total resistance: R_{\text{total}} = r + R = 20\, \Omega
X_L = \omega L = 100\pi \times 0.1 \approx 31.4\, \Omega
Total impedance: Z = \sqrt{20^2 + 31.4^2} \approx \sqrt{400 + 986} = \sqrt{1386} \approx 37.2\, \Omega
Peak current: I_0 = \dfrac{200}{37.2} \approx 5.38\, \text{A} \Rightarrow I_{\text{rms}} = \dfrac{5.38}{\sqrt{2}} \approx 3.80\, \text{A}
Power in load = I_{\text{rms}}^2 R = (3.80)^2 \times 15 \approx 14.44 \times 15 \approx 216.6\, \text{W}

Problem 8: Transformer with Load & Efficiency

Transformer Problem

Problem: A step-down transformer (turns ratio 10:1) supplies 1 A at 24 V to a resistive load. If efficiency is 90%, find (a) primary current and (b) power loss.

Solution:
Output power: P_{\text{out}} = V_s I_s = 24 \times 1 = 24\, \text{W}
Efficiency \eta = \dfrac{P_{\text{out}}}{P_{\text{in}}} \Rightarrow P_{\text{in}} = \dfrac{24}{0.9} = 26.67\, \text{W}
Primary voltage: V_p = 10 \times 24 = 240\, \text{V}
(a) I_p = \dfrac{P_{\text{in}}}{V_p} = \dfrac{26.67}{240} \approx 0.111\, \text{A}
(b) Power loss = P_{\text{in}} - P_{\text{out}} = 26.67 - 24 = 2.67\, \text{W}

Problem 9: Time-Dependent Power & Energy

Problem: In a pure resistor (R = 50 Ω) connected to v = 300 \sin(100\pi t), find energy consumed in 1 minute.

Power energy

Solution:
V_{\text{rms}} = \dfrac{300}{\sqrt{2}} \approx 212.1\, \text{V}
Power P = \dfrac{V_{\text{rms}}^2}{R} = \dfrac{(212.1)^2}{50} \approx \dfrac{45,000}{50} = 900\, \text{W}
Energy in 60 s: E = P \times t = 900 \times 60 = 54,000\, \text{J} = 54\, \text{kJ}

Problem 10: Complex Impedance & Phasor Diagram

Phasor Diagram

Problem: A circuit has Z = (30 + j40)\, \Omega connected to 200 V RMS. Find (a) current magnitude, (b) phase angle, (c) power factor, and (d) average power.

Solution:
(a) |Z| = \sqrt{30^2 + 40^2} = 50\, \Omega \Rightarrow I_{\text{rms}} = \dfrac{200}{50} = 4\, \text{A}
(b) \phi = \tan^{-1}\left(\dfrac{40}{30}\right) = \tan^{-1}(1.333) \approx 53.1^\circ (inductive)
(c) Power factor = \cos \phi = \cos 53.1^\circ \approx 0.6
(d) P = V I \cos\phi = 200 \times 4 \times 0.6 = 480\, \text{W}

✅ Note: Complex impedance notation is common in JEE Advanced!

Final Advice for Competitive Exams

  • Always identify whether the circuit is series or parallel first.
  • In resonance problems, check if it’s series (min Z) or parallel (max Z).
  • Power is only dissipated in resistance—reactive elements store and release energy.
  • Use phasor diagrams to visualize V_R, V_L, V_C relationships.

Did we miss a tricky AC concept? Share your toughest problem in the comments—we’ll feature the solution in our next post!

 🔗Go Deeper on PhysicsQandA.com

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