🎯 Problem Types & Solutions

6 problem categories with mentor-style step-by-step solutions

🎯 How to Use This Section

Each problem type follows:

Given: What information is provided
What examiner tests: The core concept being evaluated
Concept selection: Which formula/approach to use
Solution: Step-by-step working
Shortcut insight: Faster methods for exam speed

Type 1: Direct Formula Application

Board Level Easy

Problem 1: An AC source of 220V, 50Hz is connected to a resistor of 100Ω. Calculate (a) RMS current (b) Peak current (c) Average power.

🧠 What Examiner Tests

Understanding of RMS values, peak-RMS conversion, power in resistive circuit

Step 1: Identify Given Values

V_rms = 220V (AC instruments read RMS)
f = 50Hz
R = 100Ω

Step 2: Calculate RMS Current

For pure resistor: I_rms = V_rms/R
I_rms = 220/100 = 2.2A

Step 3: Calculate Peak Current

I₀ = I_rms × √2
I₀ = 2.2 × 1.414 = 3.11A

Step 4: Calculate Average Power

For resistor, cos φ = 1
P = V_rms × I_rms = 220 × 2.2 = 484W
Or: P = I_rms² R = (2.2)² × 100 = 484W

🎯 Shortcut Insight

For pure R: P = V²/R directly = (220)²/100 = 484W

No need to calculate current separately!

❌ Common Mistake

Using V₀ = 220√2 in power formula (220V is already RMS!)

JEE Main Easy

Problem 2: A coil of inductance 0.5H is connected to 100V, 50Hz AC source. Find the inductive reactance and RMS current.

Given & Find

L = 0.5H, V_rms = 100V, f = 50Hz
Find: X_L and I_rms

Calculate X_L

X_L = 2πfL
X_L = 2π × 50 × 0.5
X_L = 2 × 3.14 × 50 × 0.5 = 157Ω

Calculate I_rms

For pure inductor: I_rms = V_rms/X_L
I_rms = 100/157 = 0.64A

🔬 Key Concept

X_L ∝ f → If frequency doubles, X_L doubles, current halves

Type 2: Conceptual Reasoning

JEE Main Moderate

Problem 3: An inductor and a bulb are connected in series to an AC source. What happens to the brightness of the bulb when an iron core is inserted into the inductor?

🧠 What Examiner Tests

Understanding that inductance increases with iron core, affecting X_L and hence current

Concept Selection

Brightness ∝ Power ∝ I²R (R of bulb)
So we need to analyze what happens to current

Analysis

When iron core is inserted:
→ Inductance L increases significantly
→ X_L = ωL increases
→ Total impedance Z increases
→ Current I = V/Z decreases

Conclusion

Since I decreases → P = I²R_bulb decreases
Brightness of bulb decreases

❌ Wrong Thinking

"Iron core increases magnetic field, so brightness increases"
No! More L → More opposition → Less current

JEE Advanced Hard

Problem 4: In a series LCR circuit, the voltage across L is 30V, across C is 30V, and across R is 40V. What is the applied voltage? Explain why V ≠ V_L + V_C + V_R.

🧠 Core Concept

Voltages add vectorially (phasors), not algebraically

Why Algebraic Addition Fails

V_L leads current by 90°
V_C lags current by 90°
V_R is in phase with current
They are not in the same direction!

Vector Addition

Net reactive voltage = V_L - V_C = 30 - 30 = 0V
(V_L and V_C cancel each other)
Applied voltage = √(V_R² + (V_L - V_C)²)
V = √(40² + 0²) = 40V

🔬 Key Insight

Even though V_L and V_C are individually 30V each, they cancel in phasor sum because they're opposite in phase. This is NOT resonance (circuit has R), but a special case where X_L happens to equal X_C at given frequency.

Type 3: Multi-Step Circuit Analysis

JEE Main Moderate

Problem 5: A series LCR circuit has R=30Ω, L=0.5H, C=20μF connected to 200V, 100Hz AC source. Calculate: (a) Impedance (b) Current (c) Power factor (d) Phase angle (e) Average power.

Step 1: Calculate X_L

X_L = 2πfL = 2π(100)(0.5)
X_L = 314Ω

Step 2: Calculate X_C

X_C = 1/(2πfC) = 1/(2π × 100 × 20×10⁻⁶)
X_C = 79.6Ω

Step 3: Calculate Impedance

Z = √[R² + (X_L - X_C)²]
Z = √[30² + (314 - 79.6)²]
Z = √[900 + (234.4)²]
Z = √[900 + 54943] = √55843
Z = 236Ω

Step 4: Calculate Current

I_rms = V_rms/Z = 200/236
I_rms = 0.85A

Step 5: Calculate Power Factor

cos φ = R/Z = 30/236
cos φ = 0.127 (lagging, since X_L > X_C)

Step 6: Calculate Phase Angle

cos φ = 0.127
φ = 82.7° (current lags voltage)

Step 7: Calculate Average Power

P = V_rms I_rms cos φ
P = 200 × 0.85 × 0.127
P = 21.6W
Or: P = I²R = (0.85)² × 30 = 21.7W

🎯 Exam Strategy

Order matters: Always X_L, X_C first → Then Z → Then I → Then power

Verification: Check P = I²R matches P = VI cos φ

Type 4: Graph-Based Problems

JEE Main Moderate

Problem 6: An oscilloscope shows a sinusoidal voltage waveform with peak-to-peak voltage of 400V and time period of 20ms. Calculate: (a) Peak voltage (b) RMS voltage (c) Frequency (d) Angular frequency.

Step 1: Peak Voltage

Peak-to-peak = 2V₀
400V = 2V₀
V₀ = 200V

Step 2: RMS Voltage

V_rms = V₀/√2 = 200/1.414
V_rms = 141.4V

Step 3: Frequency

f = 1/T = 1/(20×10⁻³)
f = 50Hz

Step 4: Angular Frequency

ω = 2πf = 2π × 50
ω = 314 rad/s

❌ Graph Reading Error

Peak-to-peak = V₀ (Wrong!)
Peak-to-peak = 2V₀ (Correct)

JEE Advanced Hard

Problem 7: Two sinusoidal waveforms V(t) and I(t) are displayed on oscilloscope. V peaks at 2ms, I peaks at 3ms. Time period is 20ms. Find (a) Phase difference (b) Type of circuit (c) Power factor.

Step 1: Time Difference

Δt = 3ms - 2ms = 1ms
(Current peak comes after voltage peak)

Step 2: Phase Difference

φ = (Δt/T) × 360°
φ = (1/20) × 360° = 18°
Current lags voltage by 18°

Step 3: Circuit Type

Current lags → Inductive circuit
Circuit has net inductance (X_L > X_C)

Step 4: Power Factor

cos φ = cos(18°)
cos φ = 0.95 (lagging)

🧠 Remember

If I lags V → Inductive (X_L > X_C)
If I leads V → Capacitive (X_C > X_L)
If in phase → Resistive or resonance

Type 5: Assertion & Reason

Format

Assertion (A): A statement
Reason (R): An explanation

Choose:

(a) Both A and R are true, R is correct explanation
(b) Both A and R are true, R is not correct explanation
(c) A is true, R is false
(d) A is false, R is true

CBSE/NEET Moderate

Problem 8:
Assertion: In a purely inductive circuit, average power consumed is zero.
Reason: Current leads voltage by 90° in inductor.

Check Assertion

P_avg = V_rms I_rms cos φ
For pure L: φ = 90° → cos 90° = 0
∴ P_avg = 0 ✓
Assertion is TRUE

Check Reason

In inductor: Voltage LEADS current by 90°
Not current leads voltage!
Reason is FALSE

Answer

(c) Assertion is true, Reason is false

❌ CIVIL Memory Aid

In Capacitor: I leads V
In Inductor (VIL): V leads I

JEE Main Moderate

Problem 9:
Assertion: At resonance in series LCR circuit, current is maximum.
Reason: At resonance, impedance is minimum and equals resistance.

Check Assertion

At resonance: X_L = X_C → Z = R (minimum)
I = V/Z → Maximum when Z is minimum ✓
Assertion is TRUE

Check Reason

At resonance: Z = √[R² + (X_L - X_C)²] = √[R² + 0] = R ✓
Z is minimum = R ✓
Reason is TRUE

Is R the correct explanation?

Yes! Reason directly explains why current is maximum
(a) Both true, R correctly explains A

Type 6: Case-Based Problems

Scenario

A power transmission line operates at 11kV, 50Hz. To reduce transmission losses, engineers want to improve power factor from 0.6 to 0.9.

CBSE/JEE Main Hard

Problem 10a: What device should be used to improve power factor and why?

Analysis

Low power factor (0.6) indicates inductive load
(Industrial motors, transformers cause lagging current)

Solution

Connect capacitor in parallel

Why?
→ Capacitor draws leading current
→ Compensates lagging current from inductive load
→ Net phase angle decreases
→ cos φ increases

🔬 Practical Importance

Better power factor → Lower current for same power
→ Less I²R losses → More efficient transmission

JEE Main Hard

Problem 10b: If load draws 100A at 0.6 power factor, what current will it draw at 0.9 power factor for same power?

Concept

Real power P remains same
P = VI cos φ = constant

Before Improvement

P = V × 100 × 0.6 = 60V

After Improvement

P = V × I' × 0.9
60V = V × I' × 0.9
I' = 60V/(0.9V) = 66.7A

Answer

Current reduces from 100A to 66.7A
Reduction = 33.3%
Loss reduction = (100² - 66.7²)/100² = 55% lower!

🎯 Key Formula

For constant power: I₁ cos φ₁ = I₂ cos φ₂

I₂ = I₁ (cos φ₁)/(cos φ₂)

🎯 Problem-Solving Checklist

✓ Read carefully: Is given value peak or RMS?
✓ Check frequency: X_L and X_C depend on it
✓ Identify circuit type: Series/parallel, elements present
✓ Use correct formula: Z = √[R² + (X_L - X_C)²] not algebraic sum
✓ Power calculations: Always use P = VI cos φ
✓ Units check: Verify dimensional correctness
✓ Phase interpretation: Leading/lagging determines circuit nature