NEET Class neet Answered

⭕Help me in this question...Thanks☺️ ↪️Refer to attachment ↩️

Asked by jhajuhi19 | 11 Sep, 2019, 09:31: AM

Expert Answer

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

Answered by Thiyagarajan K | 11 Sep, 2019, 01:38: PM

Application Videos