A resistor {R_1} dissipates the power P when | vedclass

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(A) Let the voltage of the generator be $V$. Initially,the power dissipated by ${R_1}$ is given by $P = \frac{V^2}{R_1}$.When ${R_2}$ is connected in

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Decreases

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(A) Let the voltage of the generator be $V$. Initially,the power dissipated by ${R_1}$ is given by $P = \frac{V^2}{R_1}$.When ${R_2}$ is connected in series with ${R_1}$,the total resistance of the circuit becomes $R_{eq} = R_1 + R_2$.The current flowing through the circuit is $I = \frac{V}{R_1 + R_2}$.The power dissipated by ${R_1}$ in the new configuration is $P' = I^2 R_1 = \left( \frac{V}{R_1 + R_2} \right)^2 R_1 = \frac{V^2 R_1}{(R_1 + R_2)^2}$.Comparing $P'$ with $P$,we see that $P' = P \left( \frac{R_1}{R_1 + R_2} \right)^2$. Since $\frac{R_1}{R_1 + R_2} Therefore,the power dissipated by ${R_1}$ decreases.

$A$ resistor ${R_1}$ dissipates the power $P$ when connected to a certain generator. If the resistor ${R_2}$ is put in series with ${R_1}$,the power dissipated by ${R_1}$

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