Question

please solve this question with explanation...​

Answer 1

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★ Branch currents I1 , I2 , I3 , I4 and I5 are marked in figure.

★ At node A, we have, I1 = I2 + I3 .................... (1)

★ At node B, we have, I5 = I2 + I4 ........................(2)

★ At node C, we have, I3 + I5 = I1 + I4 ........................(3)

★ Kirchoff's voltage law applied to loop E1-P-A-C-Q-E1 , I1 + I3 = 20 .....................(4)

★ Kirchoff's voltage law applied to loop E2-R-B-C-S-E2 , I4 + 2I5 = 20 .....................(5)

★ Kirchoff's voltage law applied to loop A-B-C-A , 0.5 I2 + 2 I5 = I3 .....................(6)

★By solving these equations we get all branch currents.

consider eqn.(1) and (3)

I1 - I3 = I2

I1 + I3 = 20

★ By adding above eqns., we get

I1 = 10 + 0.5 I2 .....................(7)

I3 = 10 - 0.5 I2 .......................(8)

★ By sunstituting 0.5 I2 from eqn.(7) and by substituting i3 from eqn.(3), eqn.(6) is written as

I1 -10 + 2 I5 = I4 + I1 - I5 or 3 I5 - I4 = 10 ..................(9)

★ Eqn.(5) and eqn.(9) are two linear eqns. in I4 and I5 , By solving we get

I4 = 8 A , I5 = 6 A

using eqn.(2), I2 = I5 - I4 = 6 - 8 = -2A

★ By substituting I2 value in eqn.(7) and (8), we get

I1 = 10 -1 = 9 A

I3 = 10+1 = 11 A

★ Branch currents are, I1 = 9 A, I2 = -2 A , I3 = 11 A , I4 = 8 A and I5 = 6 A

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