posted by Anonymous .
A 65.0-Ù resistor is connected in parallel with a 117.0-Ù resistor. This parallel group is connected in series with a 22.0-Ù resistor. The total combination is connected across a 15.0-V battery
Find the current in the 117.0-Ù resistor.
Find the power dissipated in the 117.0-Ù resistor
total resistance parallel branch:
Req=65*117/(182)= 41.8 ohms
total circuit resistance: 41.8+22=63.8 ohms
total current: 15/63.8=.235a
voltage across the parallel branch:
current in 117 = 9.82/117 amps