Q. 93.9( 49 Votes )

Determine the current in each branch of the network shown in Fig. 3.30:


Answer :

The current through the labelled diagram shows the current flowing the respective branches:


For the closed circuit ABDA, potential is zero i.e.,


10I2 + 5I4 − 5I3 = 0


2I2 + I4 −I3 = 0


I3 = 2I2 + I4 ... (1)


For the closed circuit BCDB, 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 circuit ABCFEA, 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) in equation (1), we obtain


I3 = 2I2 + I4


− 4I4 = 2I2


I2 = − 2I4 ... (5)


It is evident from the given figure that,


I1 = I3 + I2 ... (6)


Putting equation (6) in equation (3), 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


I4 = -2/17Ampere


Equation (4) reduces to


I3 = − 3(I4)


I3 = -3 × -2/17


I3 = 6/17Ampere


I2 = -2(I4)


I2 = -2 × -2/17


I2 = 4/17Ampere


I2 - I4 = 6/17Ampere


I3 + I4 = 6/17Ampere


I1 = I2 + I3 = 4/17 + 6/17 = 10/17 Ampere


Therefore, current in branch AB = 4/17Ampere


In branch BC = 6/17Ampere


In branch CD = -4/17Ampere


In branch AD = 6/17Ampere


In branch BD = -2/17Ampere


Total current = 4/17 + 6/17 + -4/17 + 6/17 + -2/17 = 10/17Ampere.

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