Question:

Consider a simple electric circuit with three resistors, R1, R2, and R3, connected in series to a battery with emf (ε) and internal resistance (r) as shown below:

The resistors have the following values: R1 = 4 Ω, R2 = 6 Ω, and R3 = 8 Ω. The battery has an emf of 12 V and an internal resistance of 2 Ω.

a) Calculate the equivalent resistance (Req) of the circuit.

b) Determine the total current (I) flowing through the circuit.

c) Find the potential difference across each resistor (V1, V2, and V3).

d) Calculate the power dissipated by each resistor (P1, P2, and P3).

Answer:

a) To calculate the equivalent resistance (Req) of the circuit, we need to add the resistances of all the resistors in series:

Req = R1 + R2 + R3

Substituting the given values:

Req = 4 Ω + 6 Ω + 8 Ω Req = 18 Ω

b) To determine the total current (I) flowing through the circuit, we can use Ohm's Law:

I = ε / (Req + r)

Substituting the given values:

I = 12V / (18Ω + 2Ω) I = 12V / 20Ω I = 0.6 A

c) To find the potential difference across each resistor (V1, V2, and V3), we can use Ohm's Law again:

V1 = I * R1 V1 = 0.6 A * 4 Ω V1 = 2.4 V

V2 = I * R2 V2 = 0.6 A * 6 Ω V2 = 3.6 V

V3 = I * R3 V3 = 0.6 A * 8 Ω V3 = 4.8 V

d) To calculate the power dissipated by each resistor (P1, P2, and P3), we can use the formula:

P = V * I

P1 = V1 * I P1 = 2.4 V * 0.6 A P1 = 1.44 W

P2 = V2 * I P2 = 3.6 V * 0.6 A P2 = 2.16 W

P3 = V3 * I P3 = 4.8 V * 0.6 A P3 = 2.88 W

Therefore, the answers are as follows:

a) Req = 18 Ω b) I = 0.6 A c) V1 = 2.4 V, V2 = 3.6 V, V3 = 4.8 V d) P1 = 1.44 W, P2 = 2.16 W, P3 = 2.88 W