Step 1: Read the circuit:
The 6 V cell is joined across the left corner and the right corner of the bridge.
The arms are: left-top \(20\,\Omega\), top-right \(2\,\Omega\), left-bottom \(12\,\Omega\) and bottom-right \(1\,\Omega\).
The \(3\,\Omega\) resistor joins the top corner T to the bottom corner B.
Step 2: Check the balance condition:
A balanced bridge needs \(\frac{20}{2}=\frac{12}{1}\), that is \(10=12\). This is false.
So the bridge is not balanced and a current flows through the \(3\,\Omega\) branch.
The \(3\,\Omega\) resistor also loads the bridge, so we must treat the whole network.
Step 3: Apply the junction rule at T and B:
Take the left corner at \(6\,\text{V}\) and the right corner at \(0\,\text{V}\). Let the potentials of T and B be \(V_T\) and \(V_B\).
At T, current in from the left arm equals current out through the right arm and the \(3\,\Omega\) branch:
\[
\frac{6-V_T}{20}=\frac{V_T}{2}+\frac{V_T-V_B}{3}
\]
At B, current in from the left arm and from the \(3\,\Omega\) branch equals current out through the right arm:
\[
\frac{6-V_B}{12}+\frac{V_T-V_B}{3}=\frac{V_B}{1}
\]
Step 4: Simplify the two equations:
Multiply the first by 60 and the second by 12.
\[
53V_T-20V_B=18
\]
\[
4V_T-17V_B=-6
\]
Step 5: Solve for the potentials:
From the second equation \(V_B=\frac{6+4V_T}{17}\). Put this in the first equation.
\[
901V_T-120-80V_T=306 \;\Rightarrow\; 821V_T=426
\]
\[
V_T=\frac{426}{821}\,\text{V}\approx 0.519\,\text{V},\qquad V_B=\frac{390}{821}\,\text{V}\approx 0.475\,\text{V}
\]
Step 6: Find the current in the \(3\,\Omega\) branch:
\[
V_T-V_B=\frac{36}{821}\,\text{V}
\]
\[
I=\frac{V_T-V_B}{3}=\frac{12}{821}\,\text{A}\approx 1.46\times10^{-2}\,\text{A}
\]
This is about 14.6 mA. It flows from T to B when the left corner is the positive terminal.
Step 7: Check against a common mistake:
If we leave out the loading by the \(3\,\Omega\) resistor, we get \(\frac{12}{143\times 3}\approx 2.8\times10^{-2}\,\text{A}\). That value is wrong, because the \(3\,\Omega\) branch itself changes \(V_T\) and \(V_B\).
Final Answer:
\[
\boxed{I_{3\,\Omega}=\frac{12}{821}\,\text{A}\approx 1.46\times10^{-2}\,\text{A}}
\]
