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Q.
Resistance of conductor is doubled keeping the potential difference across it constant. The rate of generation of heat will
Current Electricity
Solution:
We know that heat generated in a current carrying conductor is expressed as: $H = I ^{2} R t$
Now, $V=I R \Rightarrow I=\frac{V}{R}$
So, $H=\frac{V^{2}}{R^{2}} \times R \times t$
or, $H=\frac{V^{2} \cdot t}{R}$
or, $\frac{H}{t}=\frac{V^{2}}{t}=$ rate of heat generation
Now, as par given condition, V remains unchanged.
So, $\frac{ H }{ t } \propto \frac{1}{ R }$
This means, when resistance is doubled, rate of heat generation will be halved.