The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become(A) two times. (B) half.(C) one-fourth. (D) four times.

(A) two times. 

Explanation

We know that, the resistance is given as:

$V=IR$             

where $V$ = voltage, $I$ = current, and $R$ = Resistance.

According to the question, all the parameters of the circuit remain unchanged. Also, the voltage in the circuit remains unchanged. So, the current will be given as:

$I=\frac {V}{R}$       ..........................(1)

This equation shows that resistance is inversely proportional to the current, which means when resistance increases, current decreases.

Now, 

The heating effect due to resistance is given as,

$H=I^2Rt$

From equation (1) it can be expanded in terms of voltage as-

$H=\frac {V^2}{R}t$

This shows that $H$ is inversely proportional to $R$ $(H\propto\frac {1}{R})$.

Let, the initial value of heat is ${H_1}$, and the value after reducing the resistance is ${H_2}$.

Also, let the initial value of resistance is ${R_1}$ and the value after reducing the resistance to half is ${R_2}$ $(R_2=\frac {1}{2}R_1)$.

Thus, we have the equation-

$\frac {H_1}{H_2}=\frac {R_2}{R_1}$

$H_2=\frac {R_1}{0.5R_1}{H_1}$    $(\because R_2=\frac {1}{2}R_1=0.5R_1)$

$H_2=\frac {10}{5}{H_1}$

$H_2=2{H_1}$

Thus, when the resistance is reduced to half the initial value, the heating effect in the resistor will become two times.