Current Electricity

JEE Main level, numerical answer, medium

10 of 380

Question 10 of 380

The current flowing through a conductor connected across a source is 2A2A and 1.2A1.2A at 0oC{0}^{o}C and 100oC{100}^{o}C respectively. The current flowing through the conductor at 50oC{50}^{o}C will be _____ ×102mA\times {10}^{2}mA.

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Answer: 15

By Ohm's law, V=IRV=IR.
Let the current and resistance at 0∘C0^{\circ}C be {i}_{0}&{R}_{0} respectively.
Let the current and resistance at 100∘C100^{\circ}C be {i}_{100}&{R}_{100} respectively.
By the given condition,
i0R0=i100R100⇒2R0=1.2R0(1+100α)⇒1+100α=53⇒50α=13{i}_{0}{R}_{0}={i}_{100}{R}_{100} \Rightarrow 2{R}_{0}=1.2{R}_{0}(1+100\alpha ) \Rightarrow 1+100\alpha =\dfrac{5}{3} \Rightarrow 50\alpha =\dfrac{1}{3}
Since the voltage remains constant,
i50R50=i0R0⇒i50=i0R0R50=2×R0R0(1+50α)=21+13=1.5A⇒i50=15×102mA{i}_{50}{R}_{50}={i}_{0}{R}_{0} \Rightarrow {i}_{50}=\dfrac{{i}_{0}{R}_{0}}{{R}_{50}}=\dfrac{2\times {R}_{0}}{{R}_{0}\left(1+50\alpha \right)}=\dfrac{2}{1+\dfrac{1}{3}}=1.5A \Rightarrow {i}_{50}=15\times {10}^{2}mA

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