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MCQMediumJEE 2024Combination of Resistors

JEE Physics 2024 Question with Solution

To determine the resistance RR of a wire, a circuit is designed. The value of RR is:

  • A

    1500 Ω1500 \, \Omega

  • B

    2000 Ω2000 \, \Omega

  • C

    2500 Ω2500 \, \Omega

  • D

    3000 Ω3000 \, \Omega

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A 10 kΩ10 \, \text{k}\Omega resistor is in parallel with resistance RR. From the graph, at 8 V8 \, \text{V}, the current is 4 mA=0.004 A4 \, \text{mA} = 0.004 \, \text{A}.

Find: The value of RR.

Use Ohm's law for the equivalent resistance:

Rt=VI=80.004=2000 ΩR_t = \frac{V}{I} = \frac{8}{0.004} = 2000 \, \Omega

Since RR is in parallel with 10000 Ω10000 \, \Omega,

1Rt=1R+110000\frac{1}{R_t} = \frac{1}{R} + \frac{1}{10000}

Substituting Rt=2000 ΩR_t = 2000 \, \Omega,

12000=1R+110000\frac{1}{2000} = \frac{1}{R} + \frac{1}{10000}

Now solve for RR:

1R=12000−110000=510000−110000=410000\frac{1}{R} = \frac{1}{2000} - \frac{1}{10000} = \frac{5}{10000} - \frac{1}{10000} = \frac{4}{10000} R=100004=2500 ΩR = \frac{10000}{4} = 2500 \, \Omega

Therefore, the resistance is 2500 Ω2500 \, \Omega. The correct option is C.

Equivalent Resistance Formula

Given: The equivalent resistance of RR in parallel with 104 Ω10^4 \, \Omega is obtained from the given voltage-current pair E=4 VE = 4 \, \text{V} and I=2 mAI = 2 \, \text{mA}.

Find: The value of RR.

For two resistors in parallel,

Req=104R104+RR_{\text{eq}} = \frac{10^4 R}{10^4 + R}

Using Ohm's law,

I=EReqI = \frac{E}{R_{\text{eq}}}

So,

2×10−3=4104R104+R2 \times 10^{-3} = \frac{4}{\frac{10^4 R}{10^4 + R}}

Simplifying,

2×10−3=4(104+R)104R2 \times 10^{-3} = \frac{4(10^4 + R)}{10^4 R} 2×104R=4(104+R)2 \times 10^4 R = 4(10^4 + R) 20R=40000+4R20R = 40000 + 4R 16R=4000016R = 40000 R=4000016=2500 ΩR = \frac{40000}{16} = 2500 \, \Omega

Therefore, the resistance is 2500 Ω2500 \, \Omega. This shortcut works because the graph first gives the equivalent resistance, after which the parallel combination formula directly yields RR.

Common mistakes

  • Using series combination instead of parallel combination is incorrect because the circuit places RR and 10 kΩ10 \, \text{k}\Omega across the same potential difference. Use the parallel resistance relation, not simple addition.

  • Substituting 4 mA4 \, \text{mA} directly as 4 A4 \, \text{A} is wrong because milliampere must be converted to ampere. Always write 4 mA=0.004 A4 \, \text{mA} = 0.004 \, \text{A} before applying Ohm's law.

  • Taking RtR_t as the required resistor RR is incorrect because RtR_t is the equivalent resistance of the parallel combination. First find RtR_t from V/IV/I, then use the parallel formula to obtain RR.

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