In the circuit shown, the value of currents $I_{1}, I_{2}$ and $I_{3}$ are

3 A, \frac{- 3}{2} A, \frac{9}{2} A

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\frac{9}{2} A, 3 A, \frac{- 3}{2} A

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5 A, 4 A, - 3 A

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7 A, \frac{5}{4} A, \frac{9}{2} A

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Solution

The correct option is A $3 A, \frac{- 3}{2} A, \frac{9}{2} A$
Applying Kirchhoff;s voltage law,
In loop $I, - 27 - 6 I_{2} - 2 I_{1} + 24 = 0$
$6 I_{2} + 2 I_{1} = - 3 . . . . (i)$
In loop $I I, - 27 - 6 I_{2} + 4 I_{3} = 0$
$6 I_{2} - 4 I_{3} = - 27 . . . . (i i)$
At junction $P, I_{1} - I_{2} - I_{3} = 0 . . . . (i i i)$
Solving equations (i), (ii) and (iii) we get
$I_{1} = 3 A, I_{2} = - 3 / 2 A, I_{3} = 9 / 2 A$ .

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