In the circuit shown in the figure, the current $I$ is:

6 A

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2 A

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4 A

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7 A

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Solution

The correct option is A $4 A$
Let the voltage at point $P$ be $V$ .

Hence, by Kirchoff's junction law,

$\frac{24 - V}{3} = \frac{V - 10}{2} + \frac{V - 9}{1}$

$48 - 2 V = 3 V - 30 + 6 V - 54$

$V = 12 V o l t s$

Hence, $I = \frac{24 - 12}{3} A = 4 A$

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