The figure shows an intrinsic semiconductor S | vedclass

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(B) In a semiconductor,the number of charge carriers increases with an increase in temperature,which leads to a decrease in its resistance. When the r

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With a decrease in the value of $R$

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(B) In a semiconductor,the number of charge carriers increases with an increase in temperature,which leads to a decrease in its resistance. When the resistance $R$ in the circuit is decreased,the total resistance of the circuit decreases,causing the current $I$ to increase. According to Joule's law of heating,the heat generated in the semiconductor is given by $H = I^2 R_s t$,where $R_s$ is the resistance of the semiconductor. As the current $I$ increases,the power dissipation $(I^2 R_s)$ in the semiconductor increases,which leads to an increase in the temperature of the semiconductor $S$. Thus,the semiconductor $S$ gets heated when the value of $R$ is decreased.

The figure shows an intrinsic semiconductor $S$ connected in series with a variable resistor $R$ and a constant voltage source $V$. The value of $R$ can be adjusted to maintain a constant current in the ammeter $A$. Under what condition will the semiconductor $S$ get heated? Give reasons.

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