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As the switch S is closed in the circuit shown in figure, current passed through it is

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3.0 A

6.0 A

Zero

4.5 A

answer is A.

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Detailed Solution

Let V be the potential of the junction as shown in figure. Applying junction law, we have

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$or\,\,\frac{{20 - V}}{2} + \frac{{5 - V}}{4} = \frac{{V - 0}}{2}$

or 40 – 2V + 5 – V = 2V or 5V = 45 $\Rightarrow$ V = 9V

$\therefore\,{i_3} = \frac{V}{2} = 4.5A$

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