Question

In the circuit below the voltage supplied by the battery is 9.0 V and the resistance of the resistors, R1, R2 and R3 are 2.00 Ω, 10.0 Ω and 23.0 Ω respectively, what is the power output of the resistor R1? R1 R2 Rs 2.01 W

Answer 1

R2 and R3 are in parallel

R = R2R3 / (R2+ R3) = 10*23/(10 + 23) = 6.97 ohm

R1 and R are in series

R(equi) = R1 + R = 2 + 6.97 = 8.97 ohm

I = V/R(equi)

= 9 / 8.97 = 1A

power output of R1 = I^2*R1 = 1^2*2 = 2 W

current through R2 = 1*R3 / (R2+ R3) = 1*23 / (10 + 23) = 0.7 A

current through R3 = 1*10 / (10 + 23) = 0.3 A

Power through R2 = I^2*R2 = 0.7^2*10 = 4.89 W

Power through R3 = I^2*R3 = 0.3^2*23 = 2.07 W