Question:

A circuit consists of three identical resistors connected in series, a power source of 12V, and a switch. The switch is initially open. When the switch is closed, a current of 2.0A is measured in the circuit.

a) Calculate the resistance of each resistor.

b) Determine the total resistance of the circuit.

c) Calculate the power dissipated by each resistor.

d) Calculate the total power consumed in the circuit.

Answer:

a) To calculate the resistance of each resistor, we can use Ohm's Law, which states that the resistance (R) is equal to the voltage (V) divided by the current (I):

R = V / I

In this case, the voltage is 12V and the current is 2.0A. Thus, the resistance of each resistor can be calculated as:

R = 12V / 2.0A = 6.0 Ω

Therefore, each resistor has a resistance of 6.0 Ω.

b) The total resistance of resistors connected in series can be calculated by simply adding the individual resistances together. Since we have three identical resistors of 6.0 Ω each, the total resistance of the circuit can be calculated as:

R_total = 6.0 Ω + 6.0 Ω + 6.0 Ω = 18.0 Ω

Therefore, the total resistance of the circuit is 18.0 Ω.

c) The power dissipated by a resistor can be calculated using the formula:

P = I^2 * R

where P is the power, I is the current, and R is the resistance. Since each resistor has a resistance of 6.0 Ω and the current flowing through them is 2.0A, we can calculate the power dissipated by each resistor as:

P = (2.0A)^2 * 6.0 Ω = 24.0W

Therefore, each resistor dissipates 24.0W of power.

d) The total power consumed in the circuit can be calculated by summing up the power dissipated by each resistor. Since we have three identical resistors, each dissipating 24.0W, the total power consumed is:

P_total = 24.0W + 24.0W + 24.0W = 72.0W

Therefore, the total power consumed in the circuit is 72.0W.