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The current sensitivity of a moving coil...

The current sensitivity of a moving coil galvanometer increases by `20%` when its resistance is increased by a factor 2. Calculate by what factor does the voltage sensitivity change?

Text Solution

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The correct Answer is:
40
Given, `I._s=I_s+200/(100)I_s=120/100I_s" " R.2R`
Then, initial voltage sensitivity, `V_s = I_s/R`
Now voltage sensitivity,
`V._s=(I._s)/(R.)=(120/100I_s)xx1/(2R)=3/5V_s`
`:.` % decrease in voltage sensitivity
`=(V_s-V._s)/(V_s)xx100=(V_s-3/5V_s)/(V_s)xx100=40%`
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