Question

Determine the current in each branch of the network shown in fig 3.30:

Answer 1

Current flowing throughvarious branches of the circuit is represented in the given figure.

I₁ =Current flowing through the outer circuit

I₂ =Current flowing through branch AB

I₃ =Current flowing through branch AD

I₂ −I₄ = Current flowing through branch BC

I₃ +I₄ = Current flowing through branch CD

I₄ =Current flowing through branch BD

For the closed circuitABDA, potential is zero i.e.,

10I₂+ 5I₄ − 5I₃ = 0

2I₂ +I₄ −I₃ = 0

I₃ =2I₂ + I₄ … (1)

For the closed circuitBCDB, potential is zero i.e.,

5(I₂− I₄) − 10(I₃ +I₄) − 5I₄ = 0

5I₂ +5I₄ − 10I₃ − 10I₄− 5I₄ = 0

5I₂ −10I₃ − 20I₄ = 0

I₂ =2I₃ + 4I₄ … (2)

For the closed circuitABCFEA, potential is zero i.e.,

−10 + 10 (I₁)+ 10(I₂) + 5(I₂ − I₄)= 0

10 = 15I₂+ 10I₁ − 5I₄

3I₂ +2I₁ − I₄ = 2 …(3)

From equations (1) and(2), we obtain

I₃ =2(2I₃ + 4I₄) + I₄

I₃ =4I₃ + 8I₄ + I₄

− 3I₃= 9I₄

− 3I₄= + I₃ … (4)

Putting equation (4) inequation (1), we obtain

I₃ =2I₂ + I₄

− 4I₄= 2I₂

I₂ =− 2I₄ … (5)

It is evident from thegiven figure that,

I₁ =I₃ + I₂ …(6)

Putting equation (6) inequation (1), we obtain

3I₂+2(I₃ + I₂) − I₄= 2

5I₂ +2I₃ − I₄ = 2 …(7)

Putting equations (4)and (5) in equation (7), we obtain

5(−2 I₄)+ 2(− 3 I₄) ₋ I₄= 2

− 10I₄− 6I₄ − I₄ = 2

17I₄= − 2

Equation (4) reduces to

I₃ =− 3(I₄)

Therefore, current inbranch

In branch BC =

In branch CD =

In branch AD

In branch BD =

Total current at c = $(I_2-I_4)+(I_3+I_4)=\left(\frac{6}{17}\right)+\left(\frac{4}{17}\right)=\frac{10}{17}$