Compute the equivalent resistance of the network in the figure below, and find the current in each resistor. (R1 = 3.00 Ω, R2 = 6.00 Ω, R3 = 10.0 Ω, R4 = 2.00 Ω, and ε = 55.0 V.) The battery has negligible internal resistance.
1 Answer

R12 = R1 R2 / (R1 + R2)
R12 = 3 x 6 / (3 + 6) = 18 / 9 = 2 Ω
R34 = R3 R4 / (R3 + R4)
R34 = 10 x 2 / (10 + 2) = 20 / 12 = 1.667 Ω
R = R12 + R34 = 2 + 1.667 = 3.667 Ω
I = ε / R = 55 / 3.667 = 13.636 A
V12 = I R12 = 13.636 x 2 = 27.272 V
V34 = I R34 = 13.636 x 1.667 = 22.731 V
I1 = V12 / R1 = 9.091 A
I2 = V12 / R2 = 4.545 A
I3 = V34 / R3 = 2.273 A
I4 = V34 / R4 = 11.366 A