Posted by ali-1 on .
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.
(a) Find the current in the 116.0-Ω resistor.
(b) Find the power dissipated in the 116.0-Ω resistor.
The resistors in parallel have a combined resistance R given by the formula
1/R = 1/59 + 1/116 = 0.02557
R = 39.1 ohms
Total circuit resistance = 15.0 + 39.1 = 54.1 ohms
39.1/54.1 is the fraction of the circuit voltage that is applied to the 116 ohm resistor.
Take it from there