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Current Electricity question

2020 · 8 Jan · Shift 2 · Q52
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Current Electricity question

2020 · 8 Jan · Shift 2 · Q52

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A galvanometer having a coil resistance 100 Ω\OmegaΩ gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?
  1. A
    8.9 k Ω\OmegaΩ
  2. B
    10 k Ω\OmegaΩ
  3. C
    9.9 k Ω\OmegaΩ
  4. D
    7.9 k Ω\OmegaΩ
View written solutionFree

Correct answer: C

  1. Given data

    • Galvanometer resistance: G=100 ΩG = 100\,\OmegaG=100Ω
    • Full scale deflection current: Ig=1 mA=10−3 AI_g = 1\text{ mA} = 10^{-3}\text{ A}Ig​=1 mA=10−3 A
    • Desired voltmeter range: V=10 VV = 10\text{ V}V=10 V

  2. Condition for converting galvanometer into voltmeter

    To make a voltmeter of range 10 V10\text{ V}10 V, we connect a high resistance RRR in series with the galvanometer such that full scale current IgI_gIg​ flows when the applied voltage is 10 V10\text{ V}10 V.

    So, V=Ig(R+G)V = I_g(R + G)V=Ig​(R+G)

  3. Substitute the values 10=10−3(R+100)10 = 10^{-3}(R + 100)10=10−3(R+100)

  4. Solve for RRR R+100=1010−3=10000R + 100 = \frac{10}{10^{-3}} = 10000R+100=10−310​=10000 R=10000−100=9900 ΩR = 10000 - 100 = 9900\,\OmegaR=10000−100=9900Ω

  5. Convert into kilo-ohms 9900 Ω=9.9 kΩ9900\,\Omega = 9.9\text{ k}\Omega9900Ω=9.9 kΩ

  6. Check options

    • A: 8.9 kΩ8.9\text{ k}\Omega8.9 kΩ
    • B: 10 kΩ10\text{ k}\Omega10 kΩ
    • C: 9.9 kΩ9.9\text{ k}\Omega9.9 kΩ ✅
    • D: 7.9 kΩ7.9\text{ k}\Omega7.9 kΩ

Therefore, the correct answer is Option C.

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