AP Physics 1 Exam Question - Ohm's Law

A circuit consists of a 10 Ω resistor and a 2 Ω resistor connected in series to a 6 V battery. Find:

1. The total resistance in the circuit.
2. The current flowing through the circuit.
3. The voltage drop across each resistor.
4. The power dissipated by each resistor.

Use the given information to solve the problem.

Answer:

1. The total resistance in the circuit can be calculated by summing the individual resistances:

   Total resistance (RT) = R1 + R2 = 10 Ω + 2 Ω = 12 Ω

   Therefore, the total resistance in the circuit is 12 Ω.

2. To find the current flowing through the circuit, we can use Ohm's Law:

   V = I * R

   Rearranging the equation to solve for current (I):

   I = V / R

   I = 6 V / 12 Ω = 0.5 A

   Therefore, the current flowing through the circuit is 0.5 A.

3. The voltage drop across each resistor can be calculated using Ohm's Law:

   V = I * R

   For the 10 Ω resistor:

   V1 = I * R1 = 0.5 A * 10 Ω = 5 V

   For the 2 Ω resistor:

   V2 = I * R2 = 0.5 A * 2 Ω = 1 V

   Therefore, the voltage drop across the 10 Ω resistor is 5 V, and the voltage drop across the 2 Ω resistor is 1 V.

4. The power dissipated by each resistor can be calculated using the formula:

   P = I^2 * R

   For the 10 Ω resistor:

   P1 = (0.5 A)^2 * 10 Ω = 0.25 * 10 = 2.5 W

   For the 2 Ω resistor:

   P2 = (0.5 A)^2 * 2 Ω = 0.25 * 2 = 0.5 W

   Therefore, the power dissipated by the 10 Ω resistor is 2.5 W, and the power dissipated by the 2 Ω resistor is 0.5 W.

This completes the solution for the given AP Physics 1 exam question.