## Answer :

**Answer:**

**5.0 Ω: I = 1.43 A, P = 10.2 W**

**9.0 Ω: I = 1.43 A, P = 18.4 W**

**30.0 Ω: I = 0.667 A, P = 13.3 W**

**18.0 Ω: I = 1.11 A, P = 22.2 W**

**14.0 Ω: I = 1.43 A, P = 28.6 W**

**80.0 Ω: I = 0.25 A, P = 5.00 W**

**Explanation:**

According to **Ohm's law**, the **voltage drop** across a resistor is equal to the **current **times the **resistance **(V = IR). Elements in parallel have the same voltage drop, and resistors in series have an **equivalent resistance **equal to their sum. The **power **dissipated by a resistor is equal to the **voltage **drop times the **current **(P = VI), or the **resistance **times the **square of the current** (P = I²R).

In this circuit, the 5.0 Ω resistor and 9.0 Ω resistor are in series. Together, they are in parallel with the battery, as are the other four resistors. Using Ohm's law, we can find the currents, then use that to find the power.

The 5.0 Ω resistor and 9.0 Ω resistor are in series. Their combined resistance is 5.0 + 9.0 = 14.0 Ω. The current in this branch is therefore:

V = IR

20.0 V = I (14.0 Ω)

I = 1.43 A

The power dissipated by the 5.0 Ω resistor is:

P = (1.43 A)² (5.0 Ω)

P = 10.2 W

The power dissipated by the 9.0 Ω resistor is:

P = (1.43 A)² (9.0 Ω)

P = 18.4 W

Next, the current through the 30.0 Ω resistor is:

V = IR

20.0 V = I (30.0 Ω)

I = 0.667 A

The power dissipated by the 30.0 Ω resistor is:

P = (0.667 A)² (30.0 Ω)

P = 13.3 W

Repeat the process for the other resistors.

The current through the 18.0 Ω resistor is:

V = IR

20.0 V = I (18.0 Ω)

I = 1.11 A

The power dissipated by the 30.0 Ω resistor is:

P = (1.11 A)² (18.0 Ω)

P = 22.2 W

The current through the 14.0 Ω resistor is:

V = IR

20.0 V = I (14.0 Ω)

I = 1.43 A

The power dissipated by the 14.0 Ω resistor is:

P = (1.43 A)² (14.0 Ω)

P = 28.6 W

The current through the 80.0 Ω resistor is:

V = IR

20.0 V = I (80.0 Ω)

I = 0.250 A

The power dissipated by the 80.0 Ω resistor is:

P = (0.25 A)² (80.0 Ω)

P = 5.00 W