AP Physics 1 Exam Question: Resistance

A current of 3.5 A flows through a circuit containing a resistor and a battery. The potential difference across the battery is 12 V. The resistor is then replaced with another resistor of the same length, but double the cross-sectional area. The new potential difference across the battery is 6 V. Determine the resistance of the original and replacement resistors.

Solution:

Let's denote the resistance of the original resistor as R1 and the resistance of the replacement resistor as R2. We can use Ohm's Law to solve for the resistance of each resistor.

Ohm's Law states that the potential difference (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R). Mathematically, it can be written as:

V = I * R

From the problem, we have the following information:

For the original resistor:

• Current (I1) = 3.5 A
• Potential difference (V1) = 12 V

For the replacement resistor:

• Current (I2) = 3.5 A
• Potential difference (V2) = 6 V

Using Ohm's Law, we can set up two equations for the original and replacement resistors:

1. For the original resistor: V1 = I1 * R1

2. For the replacement resistor: V2 = I2 * R2

Let's solve for R1 first:

From equation 1: V1 = I1 * R1 12 V = 3.5 A * R1

We can solve for R1 by rearranging the equation: R1 = V1 / I1 R1 = 12 V / 3.5 A R1 ≈ 3.43 Ω

Now, let's solve for R2:

From equation 2: V2 = I2 * R2 6 V = 3.5 A * R2

We can solve for R2 by rearranging the equation: R2 = V2 / I2 R2 = 6 V / 3.5 A R2 ≈ 1.71 Ω

The resistance of the original resistor (R1) is approximately 3.43 Ω, and the resistance of the replacement resistor (R2) is approximately 1.71 Ω.