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As the switch \mathrm{ S}  is closed in the circuit shown in figure, current passing through it is

Option: 1

4.5\mathrm{A}


Option: 2

6.0\mathrm{A}


Option: 3

3.0\mathrm{A}


Option: 4

\mathrm{Zero}


Answers (1)

Let the potential at junction \mathrm{C}, is \mathrm{V}. Applying Krchhoff’s junction law at \mathrm{C}


\text { or } \quad \frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}}{2}+\frac{\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{C}}}{4}=\frac{\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}}{2}

\text { or } \quad \frac{20-\mathrm{V}}{2}+\frac{5-\mathrm{V}}{4}=\frac{\mathrm{V}-0}{2}

\text { or } \quad 5 \mathrm{~V}=45

\therefore \quad \mathrm{V}=9 \mathrm{~V}

\therefore \quad$ Current through switch $\mathrm{S}, \mathrm{i}_3=\frac{9}{2}=4.5 \mathrm{~A}

Hence option 1 is correct.



 

Posted by

Kshitij

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