AP Physics 1 Exam Question:

A circuit consists of a 9.0V battery connected in series with a resistor, R1, and an ammeter. The ammeter measures a current of 0.80A. Another resistor, R2, is connected in parallel with R1. The potential difference across R2 is measured to be 6.0V. Determine the value of R1 and R2.

Answer:

Let's break down the given information:

• Voltage of the battery (V) = 9.0V
• Current measured by the ammeter (I_am) = 0.80A
• Potential difference across R2 (V_R2) = 6.0V

According to Ohm's law, the relationship between voltage (V), current (I), and resistance (R) can be expressed as V = I * R.

First, let's determine the resistance of R1 using the voltage and current:

V_R1 = I_am * R1

Since V_R1 = V - V_R2 (as R1 and R2 are in series), we can substitute the values:

V - V_R2 = I_am * R1

9.0V - 6.0V = 0.80A * R1

3.0V = 0.80A * R1

Next, let's determine the resistance of R2 using the potential difference and current:

I_R2 = V_R2 / R2

Substituting the values:

0.80A = 6.0V / R2

Solving for R2:

R2 = 6.0V / 0.80A = 7.5Ω

Now, we can substitute the calculated value of R2 back into the equation for R1:

3.0V = 0.80A * R1

R1 = 3.0V / 0.80A = 3.75Ω

Therefore, the value of R1 is 3.75Ω and the value of R2 is 7.5Ω.