Series in a Parallel Circuit Determine the power dissipated

Series in a Parallel Circuit Determine the power dissipated in resistor # 3. R1 17 R2 R3 15 42 6 V Enter Your Answers Below Don\'t Enter Units Submit Your Name: Power (W):

Solution

The voltage across the circuit is 6 volts.

The net resistance = 57*17/74 = 13.095 Ohms

Hence the current in the lower line = 6 / 57

and that in upper line = 6/ 17

Therefore energy dissipated in a resistor with resistance R and current I is R*I^2

That is, energy dissipated in each of the resistors is 17 ohm -> 36*17/289 = 36/17

Energy in 15 ohm resistor = 36*15/57*57 = 0.166

Energy in 42 ohm resistor = 36*42/57*57 = 0.465 units