Refer to the figure given below. The values of \(I_1\), \(I_2\) and \(I_3\) are __________.

Using the principle of superposition, we analyse the circuit with one source active at a time.

Case 1: Only the \(10\text{ V}\) source is active (the \(5\text{ V}\) source is replaced by a short circuit).

With the \(5\text{ V}\) source shorted, the top and bottom nodes of the bridge are directly connected by a wire, forming a Wheatstone bridge.

Checking for balance:
Left arm ratio: \(\dfrac{4}{4} = 1\)
Right arm ratio: \(\dfrac{2}{2} = 1\)

The ratios are equal, so the bridge is balanced. Hence, no current flows through the middle branch connecting the two nodes, and the contribution to \(I_2\) from the \(10\text{ V}\) source is \(0\text{ A}\).

Equivalent resistance of the balanced bridge:
\((4 + 2)\,\Omega \parallel (4 + 2)\,\Omega = 6\,\Omega \parallel 6\,\Omega = 3\,\Omega\)

Total resistance including the series \(1\,\Omega\) resistor:
\(R_{eq} = 3 + 1 = 4\,\Omega\)

Current drawn from the \(10\text{ V}\) source:
\(I_{1(10V)} = \dfrac{10}{4} = 2.5\text{ A}\)

This current splits equally between the two \(6\,\Omega\) branches, so the current through the bottom \(4\,\Omega\) resistor is:
\(I_{3(10V)} = \dfrac{2.5}{2} = 1.25\text{ A}\)

Case 2: Only the \(5\text{ V}\) source is active (the \(10\text{ V}\) source is replaced by a short circuit).

Now the \(5\text{ V}\) source is applied across the top and bottom nodes, and the \(1\,\Omega\) resistor lies in the middle branch between the left and centre nodes.

For the left branch (\(4\,\Omega\) and \(4\,\Omega\) in series), the potential at the midpoint is \(\dfrac{5}{2} = 2.5\text{ V}\).
For the right branch (\(2\,\Omega\) and \(2\,\Omega\) in series), the potential at the midpoint is also \(\dfrac{5}{2} = 2.5\text{ V}\).

Since both midpoints are at the same potential, no current flows through the \(1\,\Omega\) middle branch. Hence, the contribution to \(I_1\) from the \(5\text{ V}\) source is \(0\text{ A}\).

Equivalent resistance seen by the \(5\text{ V}\) source:
\((4 + 4)\,\Omega \parallel (2 + 2)\,\Omega = 8\,\Omega \parallel 4\,\Omega = \dfrac{8 \times 4}{8 + 4} = \dfrac{8}{3}\,\Omega\)

Total current supplied by the \(5\text{ V}\) source:
\(I_{2(5V)} = \dfrac{5}{8/3} = \dfrac{15}{8} = 1.875\text{ A}\)

Current through the left branch (which includes the bottom \(4\,\Omega\) resistor):
\(I_{3(5V)} = \dfrac{5}{8} = 0.625\text{ A}\)

Applying superposition:
\(I_1 = I_{1(10V)} + I_{1(5V)} = 2.5 + 0 = 2.5\text{ A}\)
\(I_2 = I_{2(10V)} + I_{2(5V)} = 0 + 1.875 = 1.875\text{ A}\)
\(I_3 = I_{3(10V)} + I_{3(5V)} = 1.25 + 0.625 = 1.875\text{ A}\)

Hence, the correct option is \((1)\ I_1 = 2.5\text{ A},\ I_2 = 1.875\text{ A},\ I_3 = 1.875\text{ A}\).