# Please answer it teacher..

Let the initial resistance of the resistor be R.

For a current I in the circuit, heat generated by the resistor over a time interval t is

$H={I}^{2}Rt$

It has been reduced to half. Thus, the new resistance will be R'=R/2.

From Ohm's law:

$V=IR\phantom{\rule{0ex}{0ex}}or,I\propto \frac{1}{R},providedVremianscons\mathrm{tan}t.$

Assuming all other parameters of the circuit remaining unchanged, decreasing the resistance to half must increase the double the current in the circuit

Now the current in the circuit would be $I\text{'}=2I$

Thus,

for the same time period the amount of heat produced across the resistor would be

$H\text{'}=I{\text{'}}^{2}R\text{'}t={\left(2I\right)}^{2}\left(\frac{R}{2}\right)t=2{I}^{2}Rt=2H\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}theheatingeffectsintheresistorincreasestwotimes.\phantom{\rule{0ex}{0ex}}Ans\left(A\right)$

Regards

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