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Current Electricity question

2019 · 11 Jan · Shift 2 · Q55
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Current Electricity question

2019 · 11 Jan · Shift 2 · Q55

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A galvanometer having a resistance of 20 Ω\OmegaΩ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:
  1. A
    120 Ω\OmegaΩ
  2. B
    125 Ω\OmegaΩ
  3. C
    80 Ω\OmegaΩ
  4. D
    100 Ω\OmegaΩ
View written solutionFree

Correct answer: C

  1. Given data
  • Galvanometer resistance: G=20 ΩG = 20\,\OmegaG=20Ω
  • Number of divisions on one side: 303030
  • Figure of merit: k=0.005 A/divisionk = 0.005\,\text{A/division}k=0.005A/division
  • Required voltmeter range: V=15 VV = 15\,\text{V}V=15V
  1. Find full-scale current of the galvanometer

Figure of merit means current required per division. So, for full-scale deflection:

Ig=30×0.005=0.15 AI_g = 30 \times 0.005 = 0.15\,\text{A}Ig​=30×0.005=0.15A

  1. Use voltmeter conversion formula

To convert a galvanometer into a voltmeter of range VVV, we connect a high resistance RRR in series such that

V=Ig(G+R)V = I_g(G+R)V=Ig​(G+R)

Substituting values:

15=0.15(20+R)15 = 0.15(20 + R)15=0.15(20+R)

  1. Solve for RRR

20+R=150.15=10020 + R = \frac{15}{0.15} = 10020+R=0.1515​=100

R=100−20=80 ΩR = 100 - 20 = 80\,\OmegaR=100−20=80Ω

  1. Check options
  • A: 120 Ω120\,\Omega120Ω ✗
  • B: 125 Ω125\,\Omega125Ω ✗
  • C: 80 Ω80\,\Omega80Ω ✓
  • D: 100 Ω100\,\Omega100Ω ✗

Therefore, the required series resistance is:

80 Ω\boxed{80\,\Omega}80Ω​

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