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The circuit shown in the figure below contains three resistors (R4, R2, and Rg) and three batteries (VA. Vg, and V). The resi
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Answer #1

R1 = 2 , R2=R3 = 4, VA = 25, VB = 15, VC=20.

Let suppose I1 current flowing from VA. we assume let I2 current flow in R3.

since R2 = R3 .Therefore I1 = 2*I2 (1)

now from KVL

VA = I1*R1 + VC + VB + I2*R2 (2)

put values we find I2 = 10/8 ohm

(power dissipation in R3) P = I22R3 = 6.25 W

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