# Determine the cur

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 BD = -2/17Ampere

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

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