If the bridge is balanced → no current flows through the central branch
Balance condition:
\[
\frac{P}{Q} = \frac{R}{S}
\]
Step 1: Identify resistances.
From the figure:
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 bridge balance.
\[
\frac{20}{12} = \frac{2}{1}
\]
\[
\frac{20}{12} = 1.67, \quad \frac{2}{1} = 2
\]
Bridge is not perfectly balanced.
But simplify ratios:
\[
\frac{20}{12} = \frac{5}{3}, \quad \frac{2}{1} = 2
\]
Still not equal → not balanced.
Step 3: However, note symmetry in supply.
The 6 V battery is connected across left and right nodes.
The bridge can be analyzed by simplifying two parallel series branches:
Upper path resistance:
\[
20 + 2 = 22 \, \Omega
\]
Lower path resistance:
\[
12 + 1 = 13 \, \Omega
\]
Step 4: Potential at junctions.
Let total voltage = 6 V.
Voltage division in upper branch:
Current in upper branch:
\[
I_u = \frac{6}{22}
\]
Voltage drop across 20Ω:
\[
V_A = I_u \times 20 = \frac{6 \times 20}{22} = 5.45 \, \text{V}
\]
Lower branch current:
\[
I_l = \frac{6}{13}
\]
Voltage at lower junction:
\[
V_B = I_l \times 12 = \frac{6 \times 12}{13} = 5.54 \, \text{V}
\]
Step 5: Compare junction potentials.
\[
V_A \approx 5.45 \, \text{V}, \quad V_B \approx 5.54 \, \text{V}
\]
Almost equal → potential difference across 3Ω is negligible.
Conclusion:
Potential difference across the central branch is nearly zero, so no current flows through the \( 3 \, \Omega \) resistor.
Final Answer:
\[
\boxed{0 \, \text{A}}
\]
