The problem asks us to find the current $i_o$ in the given circuit using source transformation. The circuit consists of resistors, independent current sources, and an independent voltage source.

1. Problem Description

The problem asks us to find the current $i_o$ in the given circuit using source transformation. The circuit consists of resistors, independent current sources, and an independent voltage source.

2. Solution Steps

Step 1: Transform the 5A current source in parallel with the 6$\Omega$ resistor into a voltage source in series with the 6$\Omega$ resistor.

$V = IR = 5 \text{ A} \times 6 \Omega = 30 \text{ V}$. The voltage source will be 30V with the positive terminal oriented such that the current source's arrow points towards it.

Step 2: Transform the 3A current source in parallel with the 4$\Omega$ resistor into a voltage source in series with the 4$\Omega$ resistor.

$V = IR = 3 \text{ A} \times 4 \Omega = 12 \text{ V}$. The voltage source will be 12V with the positive terminal oriented such that the current source's arrow points towards it.

Step 3: The circuit now consists of a loop with a 30V source, a 6$\Omega$ resistor, a 3$\Omega$ resistor, a 5V source, a 7$\Omega$ resistor, the current $i_o$, a 12V source, and a 1$\Omega$ resistor in series.

The total resistance in the loop is $6 + 3 + 7 + 1 + 4 = 21 \Omega$.

Step 4: Find the total voltage in the loop.

The 30V source, the 5V source, and the 12V source. The 30V and 12V sources drive current in the clockwise direction, while the 5V source drives current in the counter-clockwise direction. The net voltage driving current in the clockwise direction is $30 - 5 + 12 = 37 \text{ V}$.

Step 5: Calculate the current flowing in the loop. The current flows in the clockwise direction.

$I = \frac{V}{R} = \frac{37}{21} \text{ A}$.

Step 6: Relate the current flowing in the loop to the current $i_o$. The current $i_o$ flows downward through the 7$\Omega$ resistor. The loop current also flows downward through the 7$\Omega$ resistor. Therefore, $i_o = I = \frac{37}{21} \text{ A}$.

3. Final Answer

$i_o = \frac{37}{21} \text{ A} \approx 1.76 \text{ A}$