Question

Three resistors R1 = 80.9 Ω, R2 = 22.1 Ω, R3 = 70.0 Ω, and two batteries e m f1 = 40.0 V, and e m f2 = 360 V are connected as shown in the diagram below.

(a) What current flows through R1, R2, and R3?
(b) What is the absolute value of the potential difference across R1, R2, and R3?

Answer 1

Answer: the absolute value of the potential difference across R1, R2, and R3 is approximately:

V1 ≈ 187.02 V

V2 ≈ 51.05 V

V3 ≈ 161.68 V

Explanation: To find the current flowing through resistors R1, R2, and R3 and the potential difference across them, we can use Ohm's law and the rules for combining resistors in a circuit.

Part (a): Current through R1, R2, and R3:

Calculate the total resistance in the circuit (R_total):

R_total = R1 + R2 + R3

R_total = 80.9 Ω + 22.1 Ω + 70.0 Ω

R_total = 173.0 Ω

Calculate the total current (I_total) using Ohm's law:

I_total = EMF_total / R_total

For the two batteries in series, EMF_total = EMF1 + EMF2:

EMF_total = 40.0 V + 360 V

EMF_total = 400.0 V

I_total = 400.0 V / 173.0 Ω

I_total ≈ 2.312 A

Now we have the total current in the circuit, which is approximately 2.312 A.

Part (b): Potential Difference (Voltage) across R1, R2, and R3:

To find the potential difference across each resistor, we can use Ohm's law:

Voltage across R1 (V1) = I_total * R1

V1 ≈ 2.312 A * 80.9 Ω ≈ 187.02 V

Voltage across R2 (V2) = I_total * R2

V2 ≈ 2.312 A * 22.1 Ω ≈ 51.05 V

Voltage across R3 (V3) = I_total * R3

V3 ≈ 2.312 A * 70.0 Ω ≈ 161.68 V