Question

Determine the current in each branch of the network shown inFig.

Answer 1

The circuit is shown as below.

The given network has various branches with the flow of current as shown in the figure.

For the closed loop BCDB, potential is equal to zero so the expression is given as,

5( I 2 − I 4 )−10( I 3 + I 4 )−5 I 4 =0

Further simplifying the above expression, we get

5 I 2 −10 I 3 −20 I 4 =10 I 2 =2 I 3 +4 I 4 (1)

For the closed loop ABDA, potential is equal to zero so the expression is given as,

10 I 2 +5 I 4 −5 I 3 =0

Further simplifying the above expression, we get

2 I 2 + I 4 − I 3 =0 I 3 =2 I 2 + I 4 (2)

For the closed loop ABCFEA, potential is equal to zero so the expression is given as,

−10+10( I 1 )+10( I 2 )+5( I 2 − I 4 )=0

Further simplifying the above expression, we get

15 I 2 +10 I 1 −5 I 4 =10 3 I 2 +2 I 1 − I 4 =2 (3)

Where, the current in outer circuit is I 1 , the current in branch AB is I 2 , the current in branch AD is I 3 , the current in branch BD is I 4 , the current in branch CD is I 3 + I 4 and the current in branch BC is I 2 − I 4 .

By substituting the value of I 2 from equation (1) to equation (2), we get

I 3 =2( 2 I 3 +4 I 4 )+ I 4 −3 I 4 = I 3 (4)

By substituting the value of I 3 from equation (4) to equation (2), we get

−3 I 4 =2 I 2 + I 4 I 2 =−2 I 4 (5)

From the given figure the expression is given as,

I 1 = I 2 + I 3 (6)

By substituting the value of I 1 from equation (6) to equation (3), we get

3 I 2 +2( I 2 + I 3 )− I 4 =2 5 I 2 +2 I 3 − I 4 =2 (7)

By substituting the values of I 3 and I 2 from equation (4) and (5) to equation (7), we get

5( −2 I 4 )+2( −3 I 4 )− I 4 =2 I 4 =−0.118 A

By substituting the value of I 4 in equation (4), we get

I 3 =−3×( −2 17 ) =0.353 A

By substituting the value of I 4 in equation (5), we get

I 2 =−2×( −2 17 ) =0.235 A

By substituting the values of I 3 and I 2 in equation (6), we get

I 1 = 4 17 + 6 17 =0.588 A

The total current is given as,

I= I 2 +( I 2 − I 4 )+( I 3 + I 4 )+ I 3 + I 4

By substituting the given values in the above expression, we get

I= 4 17 + 6 17 + −4 17 + 6 17 + −2 17 =0.588 A

Thus, the current in outer circuit is 0.588 A, the current in branch AB is 0.235 A, the current in branch AD is 0.353 A, the current in branch BD is −0.118 A, the current in branch CD is 0.235 A, the current in branch BC is 0.353 A and the total current is 0.588 A.