Question

Determine the current in the \( 3 \, \Omega \) branch of a Wheatstone Bridge in the circuit shown in the figure.

Hint:

If Wheatstone bridge is balanced:

No current in central branch
Always check ratio of arms first

Answer 1

Step 1: Identify resistances.
From the bridge:
- Left upper arm: \( 20 \, \Omega \)
- Left lower arm: \( 12 \, \Omega \)
- Right upper arm: \( 2 \, \Omega \)
- Right lower arm: \( 1 \, \Omega \)
- Central branch: \( 3 \, \Omega \)
Step 2: Check balance condition.
\[ \frac{P}{Q} = \frac{R}{S} \implies \frac{20}{12} \stackrel{?}{=} \frac{2}{1} \]
\[ \frac{20}{12} = 1.67, \quad \frac{2}{1} = 2 \]
Not equal → bridge is not perfectly balanced.
Step 3: Analyze upper and lower branches.
- Upper branch resistance: \( 20 + 2 = 22 \, \Omega \)
- Lower branch resistance: \( 12 + 1 = 13 \, \Omega \)
Voltage supply: \( 6 \, \text{V} \)
- Current in upper branch: \( I_u = \frac{6}{22} \)
Voltage at upper junction: \( V_A = I_u \times 20 = \frac{6 \cdot 20}{22} \approx 5.45 \, \text{V} \)
- Current in lower branch: \( I_l = \frac{6}{13} \)
Voltage at lower junction: \( V_B = I_l \times 12 = \frac{6 \cdot 12}{13} \approx 5.54 \, \text{V} \)
Step 4: Compare junction potentials.
\[ V_A \approx 5.45 \, \text{V}, \quad V_B \approx 5.54 \, \text{V} \]
Potential difference across central branch is very small → negligible current.
Final Answer:
\[ \boxed{0 \, \text{A}} \]