**Circuit Analysis**

To determine the current through each of the other resistors, the voltage of the battery on the left, and the power delivered to the circuit by the battery on the right, we need to apply Ohm's Law and Kirchhoff's Laws.

**Current through Each Resistor**

Using Ohm's Law (V = I * R), we can calculate the current through each resistor.

- The current through the 4 Ω resistor is given as 3 A.
- Let's label the current through the 6 Ω resistor as I1.
- Since the 6 Ω resistor and the 9 Ω resistor are in parallel, the voltage across them is the same.
- Using Ohm's Law for the parallel resistors, we can write the equation: I1 * 6 Ω = 3 A * 9 Ω.
- Solving this equation, we find I1 = 4.5 A.
- Let's label the current through the 9 Ω resistor as I2.
- Since the 6 Ω resistor and the 9 Ω resistor are in parallel, the current through them is the same.
- Therefore, I2 = 4.5 A.

**Voltage of the Battery on the Left**

The voltage of the battery on the left can be determined using Kirchhoff's Voltage Law (KVL).

- Starting from the positive terminal of the battery, we move in a loop and encounter a drop of voltage across the 4 Ω resistor, which is given as 3 A * 4 Ω = 12 V.
- We then move across the 6 Ω resistor and the 9 Ω resistor in parallel.
- The total resistance of the parallel combination is given as 1 / (1/6 Ω + 1/9 Ω) = 3.6 Ω.
- The current passing through the parallel combination is 4.5 A, as calculated earlier.
- Therefore, the voltage drop across the parallel combination is 3.6 Ω * 4.5 A = 16.2 V.
- Finally, we move across the 8 Ω resistor, which results in a voltage drop of 3 A * 8 Ω = 24 V.
- Applying KVL, we find the voltage of the battery on the left is 12 V + 16.2 V + 24 V = 52.2 V.

**Power Delivered by the Battery on the Right**

The power delivered to the circuit by the battery on the right can be calculated using the formula P = V * I.

- The current passing through the 8 Ω resistor is 3 A.
- The voltage across the 8 Ω resistor is 3 A * 8 Ω = 24 V.
- Therefore, the power delivered by the battery on the right is 24 V * 3 A = 72 W.

In summary, the current through each of the other resistors is as follows:
- 4 Ω resistor: 3 A
- 6 Ω resistor: 4.5 A
- 9 Ω resistor: 4.5 A

The voltage of the battery on the left is 52.2 V.

The power delivered to the circuit by the battery on the right is 72 W.

Where is diagram????