Current Electricity

JEE Main level, single correct, medium

320 of 380

Question 320 of 380



Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp=1ΩR_p=1 \Omega as shown in the figure. An external resistance of Re=2ΩR_e=2 \Omega is connected via the sliding contact. The electric current in the circuit is :

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Answer: (D) 1.0 A

The circuit can be considered as



∴Req =0.5+0.5×22+0.5=(510+1025)Ω=4550=910=0.9∴i=0.90.9=1 A\begin{aligned} & \therefore \mathrm{R}_{\text {eq }}=0.5+\frac{0.5 \times 2}{2+0.5}=\left(\frac{5}{10}+\frac{10}{25}\right) \Omega \\ & =\frac{45}{50}=\frac{9}{10}=0.9 \\ & \therefore \mathrm{i}=\frac{0.9}{0.9}=1 \mathrm{~A}\end{aligned}

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