Physics Homework: Find the current in the..

Q) 57.0Ω resistor is connected in parallel with a 123.0Ω resistor. This parallel group is connected in series with a 24.0Ω resistor. The total combination is connected across a 15.0-V battery. Find
(a) the current in the 123.0Ω resistor(in A)

(b) the power dissipated in the 123.0Ω resistor(in W).

Sol:

Equivalent Resistance, R_eq= (57||123)series(24)

                          =[57*123/(57+123)]+23 = 61.95Ω

a) Total current in the combination, I= 15/61.95

= 0.242A

Voltage across parallel combination will be,

V=0.242*38.95= 9.43V

Hence current in 123ohm resistor is, I = 9.43/123= 0.0766A

b) Power dissipated, P= 9.43*0.0766 = 0.723W