Question

1. Given the circuit below, determine the current through each of the resistors and the power used by resistor 3. (20 points) R,-50Ω Battery Voltage is 120 V R2-20 Ω R, 30 2 R, 40 Ω R7-70 Ω R8-80 Ω

explain the answer a little please

Answer 1

R_1 and R_2 are in series. R_1 prime = R_1 + R_2 = 10 + 20 = 30 ohm As R_1 prime and R_3 are in parallel R_2 prime = R_1 prime times R_3/R_1 prime + R_3 = 30 times 30/30 + 30 = 900/60 = 15 ohm Now R_4 and R_6 are in series, also R_2 prime and R_5 are in series. R_3 prime = R_2 prime + R_5 = 15 + 50 = 65 ohm R_4 prime = R_4 + R_6 = 40 + 60 = 100 ohm \r\n Now R_3 prime and R_4 prime are in paralled so R_5 prime = R_3 prime times R_4 prime/R_3 prime + R_4 prime = 65 times 100/65 + 100 = 6500/165 R_5 prime = 1300/33 = 39.39 ohm R_total = R_5 prime + R_8 + R_7 = 39.39 + 80 + 70 = 189.39 ohm V = 120 v From ohm's law I = v/R = 120/189.39 = 0.6336 A T = 0.6336 tilde equal 0.63A