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Current Electricity question

2023 · 11 Apr · Shift 1 · Q66
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Current Electricity question

2023 · 11 Apr · Shift 1 · Q66

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The current sensitivity of moving coil galvanometer is increased by 25%25 \%25%. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:
  1. A
    +25%
  2. B
    −-− 50%
  3. C
    −-− 25%
  4. D
    Zero
View written solutionFree

Correct answer: A

  1. Write the relevant formulas

For a moving coil galvanometer:

  • Current sensitivity, Si=θI=NBAkS_i = \frac{\theta}{I} = \frac{NBA}{k}Si​=Iθ​=kNBA​ where NNN = number of turns, AAA = area of the coil, BBB = magnetic field, kkk = torsional constant.

  • Voltage sensitivity, Sv=θV=θ/IV/I=SiRS_v = \frac{\theta}{V} = \frac{\theta/I}{V/I} = \frac{S_i}{R}Sv​=Vθ​=V/Iθ/I​=RSi​​ where RRR is the resistance of galvanometer coil.

  1. Given change in current sensitivity

The current sensitivity is increased by 25%25\%25%.

So, Si′=1.25 SiS_i' = 1.25\,S_iSi′​=1.25Si​

  1. Resistance is kept constant

It is given that the change is made by varying only number of turns and area, while the resistance of the galvanometer coil remains constant.

Since Sv=SiRS_v = \frac{S_i}{R}Sv​=RSi​​ and RRR is constant, voltage sensitivity changes in the same ratio as current sensitivity.

Thus, Sv′=Si′R=1.25SiR=1.25SvS_v' = \frac{S_i'}{R} = \frac{1.25S_i}{R} = 1.25S_vSv′​=RSi′​​=R1.25Si​​=1.25Sv​

Hence voltage sensitivity also increases by 25%25\%25%.

  1. Check options
  • A: +25%+25\%+25% ✅
  • B: −50%-50\%−50% ❌
  • C: −25%-25\%−25% ❌
  • D: Zero ❌

Therefore, the correct option is A.

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