The given potentiometer has its wire of resistance 10 Ω. When the… | JEE Main 2021 (25 Jul, Shift 2)

Q124JEE Main 2021 · 25 JUL · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

The given potentiometer has its wire of resistance 10 $\Omega$ . When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2 $\Omega$ resistor is:

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Answer: (c)

Each half of the wire is $5\ \Omega$ ; the $2\ \Omega$ is in parallel with the half on the A side.
$5 \parallel 2 = \frac{10}{7}\ \Omega$ , total $R = \frac{10}{7} + 5 = \frac{45}{7}\ \Omega$
$I = \frac{20 \times 7}{45} = \frac{28}{9}\text{ A}$
$V_{2\Omega} = \frac{28}{9} \times \frac{10}{7} = \frac{40}{9}\text{ V}$
So the answer is (c) $\frac{40}{9}$ V.

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