Current in the \(3\,\Omega\) Branch Step 1: Identify the resistances of the bridge.
From the figure,
\[
P=20\Omega,
\qquad
Q=2\Omega,
\]
\[
R=12\Omega,
\qquad
S=1\Omega.
\]
The central branch contains
\[
3\Omega.
\]
Step 2: Check whether the bridge is balanced.
For balance,
\[
\frac{P}{Q}
=
\frac{R}{S}.
\]
Substituting values,
\[
\frac{20}{2}
=
10,
\]
\[
\frac{12}{1}
=
12.
\]
Since
\[
10\neq12,
\]
the bridge is not balanced.
Hence current flows through the \(3\Omega\) branch.
Step 3: Find potentials at the top and bottom junctions.
Let the left terminal be at \(6\) V and the right terminal at \(0\) V.
Top branch resistance:
\[
20+2=22\Omega.
\]
Current through top branch:
\[
I_t=\frac{6}{22}
=
\frac{3}{11}\,\text A.
\]
Potential at top junction:
\[
V_T
=
6-\left(\frac{3}{11}\times20\right)
\]
\[
V_T
=
6-\frac{60}{11}
=
\frac{6}{11}\,\text V.
\]
Bottom branch resistance:
\[
12+1=13\Omega.
\]
Current through bottom branch:
\[
I_b=\frac{6}{13}\,\text A.
\]
Potential at bottom junction:
\[
V_B
=
6-\left(\frac{6}{13}\times12\right)
\]
\[
V_B
=
6-\frac{72}{13}
=
\frac{6}{13}\,\text V.
\]
Step 4: Calculate current through the \(3\Omega\) resistor.
Potential difference across \(3\Omega\):
\[
V=V_T-V_B
=
\frac{6}{11}-\frac{6}{13}.
\]
\[
V
=
\frac{78-66}{143}
=
\frac{12}{143}\,\text V.
\]
Therefore,
\[
I=\frac{V}{3}
=
\frac{12}{143\times3}.
\]
\[
I
=
\frac{4}{143}\,\text A.
\]
\[
I
=
2.8\times10^{-2}\,\text A.
\]
Hence
\[
\boxed{
I_{3\Omega}
=
2.8\times10^{-2}\,\text A
}
\]
flowing from the top junction to the bottom junction.
