R1 R3 2 In the circuit in the figure, V1 s 41.45 V V2 is 45 05 V RI is 48.0 R2 i

Question

R1 R3 2 In the circuit in the figure, V1 s 41.45 V V2 is 45 05 V RI is 48.0 R2 is 91.0 and R3 is 88 0.800 W O Both batteries a e idea . what i herate at which energy s dissipated in u Submit Answer Incorrect. Tries 1/10 Previous Tries What is the rate at which energy is dissipated in R2? Submit Answer Tries 0/10 What is the rate at which energy is dissipated in R3? Submit Answer Tries 0/10 What power is emitted by battery 1? Submit Answer Tries 0/10 What power is emitted by battery 2? Submit Answer Tries 0/10

Explanation / Answer

let the current through V1 be x and through V2 be y..

So applying kirchoffs junction rule in the junction of R1,R2 and R3,

Ir1 + Ir3 = Ir2

=> Ir2 = x+y

also, applying kirchoffs loop rule to the loop containing V1, R1 and R2,

-V1 + I1 *r1 + (Ir3*r2) = 0

or -41.45 + x*48 + 91*(x+y)=0

applying kirchoffs loop rule to the loop containing V2, r3 and r2,

-45.05 + y*88 + (x+y)*91 = 0

solving the above two equations, we get

x = 0.2 A

y = 0.15 A

a)Power = i^2r

=0.2^2*48

=1.92 W

b)Power = Ir2^2*R2

=(0.2+0.15)^2*91

=11.1475 W

c)Power = Ir3^2*R3

=0.15^2*88

=1.98 W

d)Power by battery 1 = I1*V1

=0.2*41.45

=8.29 W

e)Power by battery 2 = I2*V2

=0.15*45.05

=6.7575 W

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