A 59.0-Ω resistor is connected in parallel with a 116.0-Ω resistor. This parallel group is connected in series with a 15.0-Ω resistor. The total combination is connected across a 15.0-V battery.

Question

(a) Find the current in the 116.0-Ω resistor.
______A

(b) Find the power dissipated in the 116.0-Ω resistor.
______ W

Answer 1

R1 = 59 Ohms.
R2 = 116 Ohms.
R3 = 15 Ohms

Rt=R1*R2 / (R1 + R2) + R3 = Total Res.
Rt = 59*116 / (59 + 116) + 15,

Answer 2

R1 = 59 Ohms.
R2 = 116 Ohms.
R3 = 15 Ohms

Rt=R1*R2 / (R1 + R2) + R3 = Total Res.
Rt = 59*116 / (59 + 116) + 15,
Rt = 39.1 + 15 = 54.1 Ohms.

It = I3 = Eb / Rt = 15 / 54.1 = 0.28A

V1 = V2 = Eb - I3*R3,
V1 = V2 = 15 - 0.28*15,
V1 = V2 = 15 - 4.16 = 10.84 Volts.

a. I2 = V2 / R2 = 10.84 / 116 = 0.093A

b. P2 = V2*I2 = 10.84 * 0.093 = 1.0W.