The resistance of a resistor is reduced to half of its initial value. If other parameters of the electrical circuit remain unaltered, the amount of heat produced in the resistor will become :

According to Joule's law of heating, the heat ($H$) produced in a resistor of resistance $R$ connected across a constant potential difference $V$ for a time $t$ is given by:

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

In a standard electrical circuit, the potential difference ($V$) supplied by the source (such as a battery or mains supply) is the parameter that remains unaltered.

Let the initial resistance be $R$ and the initial heat produced be $H$.

When the resistance is reduced to half, the new resistance becomes $R' = \frac{R}{2}$.

The new heat produced ($H'$) is:

H' = \frac{V^2}{R'} t = \frac{V^2}{\left(\frac{R}{2} ight)} t = 2 \left(\frac{V^2}{R} t\right) = 2H

Therefore, the amount of heat produced in the resistor becomes two times its initial value.

• Why other options are incorrect:
• Option (a) four times is incorrect because it would require the current to double while resistance remains constant, which is not the case here.
• Option (c) half is incorrect because it assumes that the current ($I$) remains constant ($H = I^2 R t$). However, when resistance is halved under a constant voltage source, the current actually doubles ($I = V/R$), which more than compensates for the reduced resistance.
• Option (d) one fourth is incorrect as it does not match the inverse relationship between resistance and heat production at a constant voltage.