**Exercise 17.44**

Part A = 82.62 dollars

Part B = 3.2 Ω

**Solutions Below:**

On average, an electric water heater operates for 1.8 h each day.

Part A

If the cost of electricity is $0.34 kWh, what is the cost of operating the heater during a 30-day month? The power of a typical electric water heater is 4500 W.

Express your answer using two decimal places.

Express your answer using two decimal places.

Since the heater operates for 1.8 hours each day and is at 4500 W, 1.8 x 30 x 4500 = 243000 W. This equals 243kW. At 0.34 each (243 * 0.34), the total cost is:

82.62 dollars

Part B

What is the resistance of a typical water heater? The current requirements of a typical electric water heater is 37.5 A.

Express your answer using two significant figures.

Express your answer using two significant figures.

The formula for power is W = I^{2} * R. Since I = 37.5 and W = 4500 (from Part A):

W = I^{2} * R

4500 = (37.5)^{2} * R

4500 = 1406.25R

R = 3.2 Ω

3.2 Ω