Question

A 61.5-Ω resistor is connected in parallel with a 146.6-Ω resistor. This parallel group is connected in series with a 16.0-Ω resistor. The total combination is connected across a 13.6-V battery. Find (a) the current and (b) the power dissipated in the 146.6-Ω resistor.

Answer 1

R1 = 61.5 Ohms.
R2 = 146.6 Ohms.
R3 = 16 Ohms.
V = 13.6 Volts.

a. Rt = R1*R2/(R1+R2) + R3
Rt = (61.5*146.6)/(61.5+146.6) + 16 =
59.32 Ohms. = Total resistance.

I = V/Rt = 13.6/59.32 = 0.229 Amps.

b. V2 = I*(R1*R2)/(R1+R2)=0.229 * 43.32 = 9.92 Volts. = Voltage across R2.

P2 = V2^2/R2 = 9.92^2/146.6=0.671 Watts.