✏️ Practice Section

Timed problems at all difficulty levels

Practice Timer

00:00

Board Level Practice (15 Problems)

🎯 Target: 3-4 minutes per problem

Focus on formula application and basic concepts

Problem 1 Easy

Calculate the RMS value of AC voltage with peak value 310V.

Answer: V_rms = V₀/√2 = 310/1.414 = 219.2V ≈ 220V

Problem 2 Easy

An inductor of 0.2H is connected to 50Hz AC source. Calculate inductive reactance.

Solution:
X_L = 2πfL = 2π(50)(0.2) = 62.83Ω
Answer: 62.83Ω

Problem 3 Easy

A capacitor of 100μF is connected to 100Hz AC. Find capacitive reactance.

Solution:
X_C = 1/(2πfC) = 1/(2π × 100 × 100×10⁻⁶)
X_C = 1/(0.0628) = 15.9Ω
Answer: 15.9Ω

Problem 4 Easy

In a pure resistor of 50Ω, RMS current is 2A. Calculate power dissipated.

Solution:
P = I_rms² R = (2)² × 50 = 200W
Answer: 200W

Problem 5 Easy

An LC circuit has L=1mH and C=10μF. Calculate resonance frequency.

Solution:
f₀ = 1/(2π√(LC))
f₀ = 1/(2π√(10⁻³ × 10⁻⁵))
f₀ = 1/(2π × 10⁻⁴) = 1592 Hz
Answer: 1.59 kHz

10 more problems available in full version...
Topics: Phase relationships, Phasor diagrams, Transformer, Power factor, RMS conversions

NEET/JEE Main Level (20 Problems)

🎯 Target: 2-3 minutes per problem

Multi-step calculations, graph reading, conceptual MCQs

Problem 1 Moderate

A series LCR circuit has R=40Ω, X_L=100Ω, X_C=60Ω. If RMS voltage is 200V, calculate: (a) Impedance (b) Current (c) Power factor

Solution:
(a) Z = √[R² + (X_L - X_C)²] = √[40² + (100-60)²] = √[1600 + 1600] = 56.6Ω
(b) I = V/Z = 200/56.6 = 3.53A
(c) cos φ = R/Z = 40/56.6 = 0.707
Answers: 56.6Ω, 3.53A, 0.707

Problem 2 Moderate

An AC source of 100V, 50Hz is connected to a coil having L=0.5H and R=10Ω. Calculate power consumed.

Solution:
X_L = 2πfL = 2π(50)(0.5) = 157Ω
Z = √(R² + X_L²) = √(100 + 24649) = 157.3Ω
I = V/Z = 100/157.3 = 0.636A
P = I²R = (0.636)² × 10 = 4.04W
Answer: 4.04W

Problem 3 Moderate

In a series LCR circuit at resonance, V_L = 200V and Q = 10. Find the applied voltage.

Solution:
Q = V_L/V at resonance
10 = 200/V
V = 200/10 = 20V
Answer: 20V

17 more problems available...
Topics: Graph analysis, Resonance, Quality factor, Power factor correction, Multi-step numericals

JEE Advanced Level (10 Problems)

🎯 Target: 5-7 minutes per problem

Multi-concept integration, complex analysis, non-standard scenarios

Problem 1 Hard

A coil of 100 turns, area 0.2m², rotates at 3000 rpm in magnetic field 0.1T. The coil has resistance 5Ω and is connected to external resistance 15Ω. Calculate: (a) Peak EMF (b) RMS power dissipated in external resistor.

Solution:
ω = 3000 rpm = 3000 × 2π/60 = 314.16 rad/s
(a) ε₀ = NBAω = 100 × 0.1 × 0.2 × 314.16 = 628.3V
(b) ε_rms = ε₀/√2 = 444.2V
I_rms = ε_rms/(R_coil + R_ext) = 444.2/20 = 22.21A
P_ext = I_rms² × R_ext = (22.21)² × 15 = 7.4kW
Answers: 628.3V, 7.4kW

Problem 2 Hard

A variable frequency AC source is connected to series LCR circuit (R=10Ω, L=0.1H, C=100μF). At what two frequencies will the power dissipated be half of maximum power?

Solution:
This requires bandwidth calculation.
f₀ = 1/(2π√(LC)) = 159.2 Hz
Q = (1/R)√(L/C) = (1/10)√(0.1/10⁻⁴) = 3.16
Δf = f₀/Q = 159.2/3.16 = 50.4 Hz
Half-power frequencies: f₁ = f₀ - Δf/2, f₂ = f₀ + Δf/2
Answers: 134 Hz and 184 Hz

8 more advanced problems available...
Topics: Complex impedance, Transient analysis, Non-sinusoidal waveforms, Coupled circuits

🎯 Practice Strategy

Start Easy: Build confidence with board-level problems
Use Timer: Develop exam speed from the beginning
No Calculator Initially: Practice mental approximations (√2≈1.4, π≈3.14)
Track Mistakes: Note which concept types you struggle with
Repeat Wrong Ones: Solve again after 2 days
Gradually Increase Difficulty: Don't jump to hard problems too soon

🔬 How Much Practice is Enough?

Minimum threshold for each level:

Board level: 95%+ accuracy, avg 3 min/problem
JEE Main: 85%+ accuracy, avg 2.5 min/problem
JEE Advanced: 70%+ accuracy, 80% attempted in time

If below threshold: Go back to concepts, don't just practice more