Answer :

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

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

10I_{2} + 5I_{4} − 5I_{3} = 0

2I_{2} + I_{4} −I_{3} = 0

I_{3} = 2I_{2} + I_{4} ... (1)

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

5(I_{2} − I_{4}) − 10(I_{3} + I_{4}) − 5I_{4} = 0

5I_{2} + 5I_{4} − 10I_{3} − 10I_{4} − 5I_{4} = 0

5I_{2} − 10I_{3} − 20I_{4} = 0

I_{2} = 2I_{3} + 4I_{4} ... (2)

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

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I_{2} + 10I_{1} − 5I_{4}

3I_{2} + 2I_{1} − I_{4} = 2 ... (3)

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

I_{3} = 2(2I_{3} + 4I_{4}) + I_{4}

I_{3} = 4I_{3} + 8I_{4} + I_{4}

− 3I_{3} = 9I_{4}

− 3I_{4} = + I_{3} ... (4)

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

I_{3} = 2I_{2} + I_{4}

− 4I_{4} = 2I_{2}

I_{2} = − 2I_{4} ... (5)

It is evident from the given figure that,

I_{1} = I_{3} + I_{2} ... (6)

Putting equation (6) in equation (3), we obtain

3I_{2} + 2(I_{3} + I_{2}) − I_{4} = 2

5I_{2} + 2I_{3} − I_{4} = 2 ... (7)

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

5(−2 I_{4}) + 2(− 3 I_{4}) − I_{4} = 2

− 10I_{4} − 6I_{4} − I_{4} = 2

17I_{4} = − 2

I_{4} = -2/17Ampere

Equation (4) reduces to

I_{3} = − 3(I_{4})

I_{3} = -3 × -2/17

I_{3} = 6/17Ampere

I_{2} = -2(I_{4})

I_{2} = -2 × -2/17

I_{2} = 4/17Ampere

I_{2} - I_{4} = 6/17Ampere

I_{3} + I_{4} = 6/17Ampere

I_{1} = I_{2} + I_{3} = 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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