Question

Question 8 5 pts The circuit shown in the figure below contains three resistors (R1, R2, and Rg) and three batteries (VA. V. and Vc). The resistor values are: Rq=2 Ohms, R2=R3-6 Ohms, and the battery voltages are VA-25V, Vo=15 V, and Vc=20V. When the circuit is connected, what will be the power dissipated by R2? Vc R R2 R3 1.0 W 20w 5.0 W 6.0W 7.5 W

Answer 1

VC I Ri Vg VA R2 b a R3 VA - 25 V, Vo = 15 V, Vc= 200, Ri=an, R2 = R3=62 Let op assume certenst I to flow in corection shower equivalent resistance of parallel combination of R₂ and R3 is, Rp= cy 1 goles Rp R2 R2 1 / 2 → Rp 1 R3 6 12 2 소 > RP= "Hico al -> Rp=30 Vc I VA Hence circuit is bous as. applying UL in closed loop, RI VB TVC - IR, -VA-IRP=0 I CRre RP2 WBA Vc - VA = I= VB + VC- VA que Ret Rp => I = (5+20-25 ) A =I10 A > Iz 2 A (2+3) Re b 5

R2 6 Vab= I x Rp > Vob@x 3) u » Vab=6U ². Power dissipated across R2, P2I Vab 2 st P2 = 6² w [P2=6W] Hence power dissipated by Re is 6w coption 4)