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Current Electricity question

2005 · Shift 0 · Q122
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Current Electricity question

2005 · Shift 0 · Q122

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A moving coil galvanometer has 150150150 equal divisions. Its current sensitivity is 101010- divisions per milliampere and voltage sensitivity is 222 divisions per millivolt. In order that each division reads 111 volt, the resistance in ohmsohmsohms needed to be connected in series with the coil will be -
  1. A
    105{10^5}105
  2. B
    103{10^3}103
  3. C
    999599959995
  4. D
    999959999599995
View written solutionFree

Correct answer: C

  1. Given data
  • Number of divisions in galvanometer scale =150=150=150
  • Current sensitivity =10=10=10 divisions per milliampere
  • Voltage sensitivity =2=2=2 divisions per millivolt
  • We want: each division should read 1 V1\text{ V}1 V

So, for full-scale deflection of 150150150 divisions, the voltmeter range should be 150 V.150\text{ V}.150 V.

  1. Find full-scale current of the galvanometer

Current sensitivity means: 10 divisions per mA10\text{ divisions per mA}10 divisions per mA

Hence, current needed for 150 divisions is Ig=15010 mA=15 mA=0.015 A.I_g=\frac{150}{10}\text{ mA}=15\text{ mA}=0.015\text{ A}.Ig​=10150​ mA=15 mA=0.015 A.

  1. Find galvanometer resistance using voltage sensitivity

Voltage sensitivity means: 2 divisions per mV2\text{ divisions per mV}2 divisions per mV

So, voltage needed for 150 divisions is Vg=1502 mV=75 mV=0.075 V.V_g=\frac{150}{2}\text{ mV}=75\text{ mV}=0.075\text{ V}.Vg​=2150​ mV=75 mV=0.075 V.

Thus galvanometer resistance is Rg=VgIg=0.0750.015=5 Ω.R_g=\frac{V_g}{I_g}=\frac{0.075}{0.015}=5\,\Omega.Rg​=Ig​Vg​​=0.0150.075​=5Ω.

  1. Convert galvanometer into voltmeter with 1 V per division

If each division corresponds to 1 V1\text{ V}1 V, then for 150 divisions full-scale voltage must be V=150 V.V=150\text{ V}.V=150 V.

For full-scale current Ig=0.015 AI_g=0.015\text{ A}Ig​=0.015 A, total resistance needed is Rtotal=VIg=1500.015=10000 Ω.R_{\text{total}}=\frac{V}{I_g}=\frac{150}{0.015}=10000\,\Omega.Rtotal​=Ig​V​=0.015150​=10000Ω.

This total includes galvanometer resistance Rg=5 ΩR_g=5\,\OmegaRg​=5Ω.

Therefore, required series resistance is R=Rtotal−Rg=10000−5=9995 Ω.R=R_{\text{total}}-R_g=10000-5=9995\,\Omega.R=Rtotal​−Rg​=10000−5=9995Ω.

  1. Check options
  • A: 105=10000010^5=100000105=100000 ❌
  • B: 103=100010^3=1000103=1000 ❌
  • C: 999599959995 ✅
  • D: 999959999599995 ❌

So the correct option is C  (9995 Ω)\boxed{C\; (9995\,\Omega)}C(9995Ω)​

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