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MCQMediumJEE 2025Kirchhoff's Laws & Circuits

JEE Physics 2025 Question with Solution

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp=1 ΩR_p = 1 \, \Omega as shown in the figure.

An external resistance of Re=2 ΩR_e = 2 \, \Omega is connected via the sliding contact.

The current ii is :

A circuit with a 0.9 V cell across the ends of a 1 ohm potentiometer wire, whose midpoint slider is connected to a 2 ohm external resistor.
  • A

    0.3 A0.3 \, \text{A}

  • B

    1.35 A1.35 \, \text{A}

  • C

    1.0 A1.0 \, \text{A}

  • D

    0.9 A0.9 \, \text{A}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The potentiometer wire has resistance Rp=1 ΩR_p = 1 \, \Omega and the sliding contact is at the middle, so each half has resistance 0.5 Ω0.5 \, \Omega. The external resistance is Re=2 ΩR_e = 2 \, \Omega and the source voltage is 0.9 V0.9 \, \text{V}.

Find: The current ii.

The circuit can be considered as

Req=0.5+0.5×22+0.5=510+1025 Ω=0.9 ΩR_{\text{eq}} = 0.5 + \frac{0.5 \times 2}{2 + 0.5} = \frac{5}{10} + \frac{10}{25} \, \Omega = 0.9 \, \Omega

Now, the current is

i=0.90.9=1 Ai = \frac{0.9}{0.9} = 1 \, \text{A}

Therefore, the current is 1.0 A1.0 \, \text{A} and the correct option is C.

Stepwise Circuit Simplification

Given: The sliding contact is at the midpoint of the potentiometer wire, so the wire is divided into two equal resistances of Rp2=0.5 Ω\dfrac{R_p}{2} = 0.5 \, \Omega each. Also, Re=2 ΩR_e = 2 \, \Omega.

Find: The current through the circuit.

Identify the combination: One half of the potentiometer wire is in series with the parallel combination of the other half and the external resistor.

So,

Req=0.5+(0.5×20.5+2)R_{\text{eq}} = 0.5 + \left(\frac{0.5 \times 2}{0.5 + 2}\right)

Now calculate the parallel part:

0.5×20.5+2=12.5=0.4 Ω\frac{0.5 \times 2}{0.5 + 2} = \frac{1}{2.5} = 0.4 \, \Omega

Hence,

Req=0.5+0.4=0.9 ΩR_{\text{eq}} = 0.5 + 0.4 = 0.9 \, \Omega

Using Ohm's law with source voltage 0.9 V0.9 \, \text{V},

i=VReq=0.90.9=1 Ai = \frac{V}{R_{\text{eq}}} = \frac{0.9}{0.9} = 1 \, \text{A}

Thus, the current is 1.0 A1.0 \, \text{A}.

The second provided approach shows a different intermediate equivalent resistance, but it still concludes the current to be 1.0 A1.0 \, \text{A}. The option consistent with the stated correct answer on the solution is C.

Common mistakes

  • Treating the full potentiometer resistance 1 Ω1 \, \Omega as a single resistor without splitting it into two halves. This is wrong because the slider is at the midpoint, creating two resistive sections of 0.5 Ω0.5 \, \Omega each. First divide the potentiometer wire correctly before simplifying the circuit.

  • Adding 0.5 Ω0.5 \, \Omega and 2 Ω2 \, \Omega directly as series resistors. This is wrong because the branch connected through the slider forms a parallel combination in the simplified network. Identify which resistors share the same two nodes before using the series or parallel formula.

  • Using Ohm's law with the wrong voltage or wrong equivalent resistance. This leads to incorrect values such as 0.3 A0.3 \, \text{A} or 0.9 A0.9 \, \text{A}. After finding ReqR_{\text{eq}}, use i=VReqi = \dfrac{V}{R_{\text{eq}}} with the source voltage 0.9 V0.9 \, \text{V}.

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