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MCQMediumJEE 2023Kirchhoff's Laws & Circuits

JEE Physics 2023 Question with Solution

A cell of emf 90 Ω90 \, \Omega is connected across a series combination of two resistors each of 100 Ω100\, \Omega resistance. A voltmeter of resistance 400 Ω400\, \Omega is used to measure the potential difference across each resistor. The reading of the voltmeter will be:

  • A

    40 V40 \, \text{V}

  • B

    45 V45 \, \text{V}

  • C

    80 V80 \, \text{V}

  • D

    90 V90 \, \text{V}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Resistance of each resistor is R=100 ΩR = 100 \, \Omega, resistance of voltmeter is Rv=400 ΩR_v = 400 \, \Omega, and emf of the cell is 90 V90 \, \text{V}.

Find: The reading of the voltmeter.

The voltmeter is connected across one resistor, so that resistor and the voltmeter are in parallel.

Req=Rv⋅RRv+R=400⋅100400+100=40000500=80 ΩR_{\text{eq}} = \frac{R_v \cdot R}{R_v + R} = \frac{400 \cdot 100}{400 + 100} = \frac{40000}{500} = 80 \, \Omega

Now this equivalent resistance is in series with the other 100 Ω100 \, \Omega resistor.

Rtotal=100+80=180 ΩR_{\text{total}} = 100 + 80 = 180 \, \Omega

Hence the circuit current is

I=EMFRtotal=90180=0.5 AI = \frac{\text{EMF}}{R_{\text{total}}} = \frac{90}{180} = 0.5 \, \text{A}

The potential difference across the resistor-voltmeter parallel combination is

V=I×Req=0.5×80=40 VV = I \times R_{\text{eq}} = 0.5 \times 80 = 40 \, \text{V}

Therefore, the reading of the voltmeter is 40 V40 \, \text{V}. The extracted solution concludes option B, but this value corresponds to option A in the listed options, so the correct defensible option is A.

Common mistakes

  • Treating the voltmeter as ideal and ignoring its finite resistance is wrong because here Rv=400 ΩR_v = 400 \, \Omega is comparable to the resistor value. First take the parallel combination of the voltmeter and the resistor across which it is connected.

  • Adding 100 Ω100 \, \Omega and 400 Ω400 \, \Omega directly is wrong because the voltmeter is connected across the resistor, not in series with it. Use the parallel formula Req=RvRRv+RR_{\text{eq}} = \frac{R_v R}{R_v + R} instead.

  • Using the total current directly with a single 100 Ω100 \, \Omega resistor to get the voltmeter reading is incorrect. The voltmeter reads the voltage across the parallel branch, so multiply the circuit current by the branch equivalent resistance 80 Ω80 \, \Omega.

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