Question

The circuit shown in the figure below contains three resistors (R1, R2, and R3) and three batteries (VA. Vg, and V). The resistor values are: R1=2 Ohms, Ry=R3=8 Ohms, and the battery voltages are VA=25 V, Vo=15 V, and Vc=20 V. When the circuit is connected, what will be the power dissipated by R3? Vc RA VA VB R2 R3

Answer 1

The R2 and R3 is in parallel combination. So the equivalent resistance will be= (8×8)/16 = 4

Applying Kirchoff's law in the circuit:

-Vc -Vb - 4i + Va - 2i =0

=> -20 -15- 4i+ 25-2i = 0

=> 6i = -10

=> i = -(10/6) Amp

The current in the circuit will be (10/6) Amp.

The voltage across the R3 resistance = (10/6)×(4)=(40/6)V

The power dissipation by R3 = (40/6)²/8 = 5.55 W

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