AP Physics 1 Exam Question: Resistance

Question:

A circuit consists of a battery with an electromotive force (emf) of 9 volts and two resistors connected in series. The first resistor, $R_1$, has a resistance of 10 ohms, while the second resistor, $R_2$, has an unknown resistance. When a current of 0.5 amperes flows through the circuit, the potential difference across $R_2$ is found to be 2 volts.

1. Calculate the resistance, $R_2$, of the second resistor.

Answer:

To calculate the resistance, $R_2$, of the second resistor, we can use Ohm's Law, which states that the potential difference across a resistor is equal to the current flowing through it multiplied by its resistance.

Ohm's Law: $V = IR$

In this case, the potential difference ($V$) across $R_2$ is 2 volts, and the current ($I$) flowing through the circuit is 0.5 amperes.

Using Ohm's Law, we can rearrange the formula to solve for resistance:

$R = \frac{V}{I}$

Now, we can substitute the given values:

$R_2 = \frac{2 \, \text{volts}}{0.5 \, \text{amperes}}$

Calculating the above expression, we find:

$R_2 = 4 \, \text{ohms}$

Therefore, the resistance of the second resistor ($R_2$) is 4 ohms.