Refer to the circuit shown. What will be the | vedclass

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(A) Let the $EMF$ of the battery be $E$. The resistors $R_2$ and $R_3$ are in parallel,so their equivalent resistance $R_p$ is:$R_p = \frac{R_2 R_3}{R

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(A) Let the $EMF$ of the battery be $E$. The resistors $R_2$ and $R_3$ are in parallel,so their equivalent resistance $R_p$ is:$R_p = \frac{R_2 R_3}{R_2 + R_3} = \frac{(4 R_1)(12 R_1)}{4 R_1 + 12 R_1} = \frac{48 R_1^2}{16 R_1} = 3 R_1$The total resistance of the circuit is $R_{	ext{net}} = R_1 + R_p = R_1 + 3 R_1 = 4 R_1$.The total current in the circuit is $I = \frac{E}{R_{	ext{net}}} = \frac{E}{4 R_1}$.The power dissipated in $R_1$ is $P = I^2 R_1 = \left(\frac{E}{4 R_1}ight)^2 R_1 = \frac{E^2}{16 R_1}$.The voltage across the parallel combination is $V_p = I R_p = \left(\frac{E}{4 R_1}ight) (3 R_1) = \frac{3 E}{4}$.The power dissipated in $R_2$ is $P_2 = \frac{V_p^2}{R_2} = \frac{(3 E / 4)^2}{4 R_1} = \frac{9 E^2 / 16}{4 R_1} = \frac{9 E^2}{64 R_1} = \frac{9}{4} \left(\frac{E^2}{16 R_1}ight) = \frac{9 P}{4}$.The power dissipated in $R_3$ is $P_3 = \frac{V_p^2}{R_3} = \frac{(3 E / 4)^2}{12 R_1} = \frac{9 E^2 / 16}{12 R_1} = \frac{3 E^2}{64 R_1} = \frac{3}{4} \left(\frac{E^2}{16 R_1}ight) = \frac{3 P}{4}$.The total power dissipation is $P_{	ext{total}} = P + P_2 + P_3 = P + \frac{9 P}{4} + \frac{3 P}{4} = P + \frac{12 P}{4} = P + 3 P = 4 P$.

Refer to the circuit shown. What will be the total power dissipation in the circuit if $P$ is the power dissipated in $R_1$ (in $P$)? It is given that $R_2 = 4 R_1$ and $R_3 = 12 R_1$.

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