If the resistance of a wire is doubled while the voltage remains constant, what happens to the current passing through it?
The current is halved
By Ohm's law, \(V = I R\), so \(I = V / R\) — at a fixed voltage, current is inversely proportional to resistance.
Let the original current be \(I_1 = V / R\). When R is doubled (R → 2R) at the same V:
\(I_2 = \dfrac{V}{2R} = \dfrac{1}{2} \cdot \dfrac{V}{R} = \dfrac{I_1}{2}\).
Hence the current is halved.
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