If a resistor is connected to a voltage source and the resistance is halved while the voltage remains constant, what happens to the heat produced in a fixed time?
The heat produced increases to twice its previous value
Joule's law of heating gives the heat produced in a resistor as \(H = I^2 R t\). When the resistor is connected across a fixed voltage source, it is more useful to express current using Ohm's law \(I = V/R\), which turns the heating expression into \(H = \dfrac{V^2}{R} \cdot t\).
Because the voltage V and the time t are held constant, the heat produced is inversely proportional to the resistance R. So if the resistance is halved, the heat produced becomes \(\dfrac{V^2}{R/2} \cdot t = 2 \cdot \dfrac{V^2}{R} \cdot t\), which is exactly twice the original heat.
Hence, the heat produced in a fixed time increases to twice its previous value.
The option saying the heat halves would only be right if one wrongly used \(H = I^2 R t\) with the current unchanged; here the current itself doubles when R halves. It cannot become four times, because that would require the current (not the voltage) to double while R stayed the same, giving \(H \propto I^2 R\). It cannot stay unchanged either, since both the current and the power dissipated clearly rise.
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