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Current Electricity question

2024 · 30 Jan · Shift 2 · Q86
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Current Electricity question

2024 · 30 Jan · Shift 2 · Q86

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
Two resistance of 100Ω100 \Omega100Ω and 200Ω200 \Omega200Ω are connected in series with a battery of 4 V4 \mathrm{~V}4 V and negligible internal resistance. A voltmeter is used to measure voltage across 100Ω100 \Omega100Ω resistance, which gives reading as 1 V1 \mathrm{~V}1 V. The resistance of voltmeter must be ‾Ω\underline{\hspace{2cm}}\Omega​Ω.
Numerical answer
View written solutionFree

Correct answer: 200

  1. Set up the circuit

The two resistors are 100 Ω100\,\Omega100Ω and 200 Ω200\,\Omega200Ω in series with a battery of 4 V4\,\text{V}4V.

A voltmeter is connected across the 100 Ω100\,\Omega100Ω resistor. Since a voltmeter has finite resistance, it will be in parallel with the 100 Ω100\,\Omega100Ω resistor.

Let the resistance of the voltmeter be RvR_vRv​.

So, the effective resistance of the parallel combination is

Rp=100Rv100+RvR_p = \frac{100R_v}{100+R_v}Rp​=100+Rv​100Rv​​

This parallel combination is in series with 200 Ω200\,\Omega200Ω.

  1. Use the given voltmeter reading

The voltmeter reads the voltage across the parallel combination, which is given as 1 V1\,\text{V}1V.

Hence, the voltage across the 200 Ω200\,\Omega200Ω resistor is

4−1=3 V4 - 1 = 3\,\text{V}4−1=3V
  1. Find the circuit current

Current through the 200 Ω200\,\Omega200Ω resistor is

I=3200 AI = \frac{3}{200}\,\text{A}I=2003​A

This is also the total current entering the parallel combination.

  1. Find the equivalent resistance of the parallel part

Since voltage across the parallel combination is 1 V1\,\text{V}1V,

Rp=VI=13/200=2003 ΩR_p = \frac{V}{I} = \frac{1}{3/200} = \frac{200}{3}\,\OmegaRp​=IV​=3/2001​=3200​Ω

So,

100Rv100+Rv=2003\frac{100R_v}{100+R_v} = \frac{200}{3}100+Rv​100Rv​​=3200​
  1. Solve for RvR_vRv​
100Rv=2003(100+Rv)100R_v = \frac{200}{3}(100+R_v)100Rv​=3200​(100+Rv​)

Multiply by 333:

300Rv=200(100+Rv)300R_v = 200(100+R_v)300Rv​=200(100+Rv​) 300Rv=20000+200Rv300R_v = 20000 + 200R_v300Rv​=20000+200Rv​ 100Rv=20000100R_v = 20000100Rv​=20000 Rv=200 ΩR_v = 200\,\OmegaRv​=200Ω
  1. Final answer

The resistance of the voltmeter is

200 Ω\boxed{200\,\Omega}200Ω​
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