Q.58 In the circuit shown below, the power dissipated across the 3 resistor is _____ W.
🔌 Circuit Understanding
The circuit consists of:
- A 30 V ideal DC source
- A 6 Ω resistor in series with the source
- A parallel combination of 6 Ω and 3 Ω
- A 2 Ω resistor in series with the rest of the circuit (return path)
All elements except the 6 Ω and 3 Ω parallel pair are in series, with the parallel branch forming a key sub-circuit.
⚙️ Step-by-Step Solution
Step 1: Parallel Branch Equivalent Resistance
Formula: \( R_p = \frac{R_1 R_2}{R_1 + R_2} \)
Here: \( R_1 = 6 \, \Omega \), \( R_2 = 3 \, \Omega \)
\( R_p = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega \)
Here: \( R_1 = 6 \, \Omega \), \( R_2 = 3 \, \Omega \)
\( R_p = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega \)
Step 2: Total Circuit Resistance
Series combination: \( R_{total} = 6 \, \Omega + 2 \, \Omega + 2 \, \Omega = 10 \, \Omega \)
Step 3: Total Current from Battery
Ohm’s Law: \( I_{total} = \frac{V}{R_{total}} = \frac{30}{10} = 3 \, A \)
Step 4: Voltage Across Parallel Branch
\( V_{parallel} = I_{total} \times R_p = 3 \, A \times 2 \, \Omega = 6 \, V \)
Step 5: Current Through 3 Ω Resistor
\( I_{3\Omega} = \frac{V_{parallel}}{3 \, \Omega} = \frac{6}{3} = 2 \, A \)
Step 6: Power Dissipation Calculation
| Method | Formula | Calculation | Result |
|---|---|---|---|
| I²R Method | \( P = I^2 R \) | \( P = (2)^2 \times 3 = 4 \times 3 \) | 12 W |
| V²/R Method | \( P = \frac{V^2}{R} \) | \( P = \frac{6^2}{3} = \frac{36}{3} \) | 12 W |
❌ Common Error Alert: The initial claim of 18 W is incorrect. Both power formulas confirm 12 W as the correct value.
🔍 Current Distribution Verification
| Branch | Resistance | Voltage | Current |
|---|---|---|---|
| 6 Ω Branch | 6 Ω | 6 V | 1 A |
| 3 Ω Branch | 3 Ω | 6 V | 2 A |
| Total | 2 Ω (equiv) | 6 V | 3 A ✓ |
Kirchhoff’s Current Law satisfied: 1 A + 2 A = 3 A
🎯 Key Learning Points for CSIR NET
- Parallel Formula: \( R_p = \frac{R_1 R_2}{R_1 + R_2} \) for two resistors
- Series Rule: Direct addition of resistances
- Ohm’s Law Priority: Total current first, then branch currents
- Power Formulas: Use \( P = I^2R \), \( P = \frac{V^2}{R} \), or \( P = VI \) based on known values
- Verification: Always cross-check with KCL and multiple power methods
📋 Quick Reference Summary
Final Answer: Power dissipated across 3 Ω resistor = 12 W
Circuit Parameters: 30 V, 10 Ω total, 3 A total current, 6 V across parallel branch
Circuit Parameters: 30 V, 10 Ω total, 3 A total current, 6 V across parallel branch
