JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Current Electricity

The given potentiometer has its wire of resistance

10Ω

. When the sliding contact is in the middle of the potentiometer wire, the potential drop across

2Ω

resistor is:

Answer: (C)

Solution

The slider at the middle divides the

10 Ω

wire into two

5 Ω

halves. The

2 Ω

resistor runs from the slider back to

A

, so it is in parallel with the half between

A

and the slider:

R_{p} = \frac{5 \times 2}{5 + 2} = \frac{10}{7} Ω

This combination is in series with the other half, so the supply sees

R = 5 + \frac{10}{7} = \frac{45}{7} Ω

and

I = \frac{20}{45/7} = \frac{28}{9} A

The whole of this current crosses the series half, and the voltage across the parallel section is the voltage across the

2 Ω

resistor:

V = I R_{p} = \frac{28}{9} \times \frac{10}{7} = \frac{40}{9} V

The potential drop across the

2 Ω

resistor is

\frac{40}{9}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option