Refer to teh circuit shown. What will be the total power dissipation in the circuit if $P$ is the power dissipated in $R_{1}$ ? It is given that $R_{2} = 4 R_{1}$ and $R_{3} = 12 R_{1}$

4 P

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7 P

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13 P

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17 P

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Solution

The correct option is A $4 P$
Total equivalent resistance= $\frac{R_{2} R_{3}}{R_{2} {+ R}_{3}} + R_{1} = \frac{48 {R_{1}}^{2}}{16 R_{1}} + R_{1} = 3 R_{1} + R_{1} = 4 R_{1}$
Power dissipation at $R_{1} = I^{2} R_{1} = P$
$∴$ Power dissipation in the total circuit= $I^{2} 4 R_{1}$
$= 4 I^{2} R_{1}$
$= 4 P$

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