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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Equivalent resistance in the second case $ = {R_1} + {R_2} = R$
Now, we know that $P \propto \frac{1}{R}$
Since in the second case the resistance $({R_1} + {R_2})$ is higher than that in the first case $(R_1).$
Therefore power dissipation in the second case will be decreased.
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