**Exercise 16.39**

Part A = 3.06 V

Part B = 3.636 * 10^{-2} Ω

Part C = 1.82 Ω

**Solutions Below:**

Part A

Express your answer in volts to three significant digits.

Electromotive force is a synonym for voltage. In this case, it’s 3.06 (the voltage of the battery, ignoring internal resistance).

3.06 V

Part B

Express your answer in ohms to four significant digits.

You know there’s a difference of 0.06 V between opening and closing the switch (3.06 – 3.00). This is due to the internal resistance of the battery. The problem also says that there’s 1.65 A of current in the circuit. Ohm’s Law gives us the formula V = IR:

V = IR

0.06 = 1.65 * R_{internal}

R_{internal} = 3.636 * 10^{-2} Ω

3.636 * 10^{-2} Ω

Part C

Express your answer in ohms to three significant digits.

Ohm’s law gives us the formula V = IR. Use 3.00 as the voltage (this is the voltage net of the battery’s internal resistance):

V = IR

3 = 1.65 * R_{circuit}

R_{circuit} = 1.82 Ω

1.82 Ω