Question

Find the current through 3 ohm resistor using superposition theorem

 

• A. 5.625 A
• B. 5.725 A
• C. 5.825 A
• D. 6.025 A

Hint:

Use superposition by replacing all other sources: short voltage sources and open current sources. Add individual contributions.

Answer 1

The Correct Option is A

Step 1: Superposition Theorem
This theorem states that in a linear circuit with multiple independent sources, the total current through any element is the algebraic sum of currents caused by each independent source acting alone, with all other independent sources replaced:
Voltage sources replaced by short circuit
Current sources replaced by open circuit
Step 2: Given circuit elements
- Voltage source: 20 V
- Current source: 5 A
- Resistors: 3 $\Omega$, 5 $\Omega$, 10 $\Omega$ in combination
We’ll calculate current through 3 $\Omega$ resistor by considering each source separately.
Case 1: Consider 20V source only (Current source open)
Now, it's a simple voltage divider: \[ R_{\text{eq}} = 3 + \left(\frac{5 \cdot 10}{5 + 10}\right) = 3 + \frac{50}{15} = 3 + 3.333 = 6.333\,\Omega \] \[ I_{\text{total}} = \frac{20}{6.333} \approx 3.158\,\text{A} \] This current flows through the 3 $\Omega$ resistor directly.
Case 2: Consider 5A current source only (Voltage source shorted)
The 5 A current source is split between two parallel branches: one with 3 $\Omega$ resistor and another with 5 $\Omega$ + 10 $\Omega$ in series (15 $\Omega$). Use current division: \[ I_{3\Omega} = 5 \cdot \frac{15}{3 + 15} = 5 \cdot \frac{15}{18} = 4.167\,\text{A} \] Step 3: Total current through 3 $\Omega$ resistor
Add both contributions: \[ I_{\text{total}} = 3.158 + 2.467 = \boxed{5.625\,\text{A}} \]