When a resistance of `100Omega` is connected in series with a galvanometeer of resistance R, its range is V. To double its range, a resistance of `1000Omega` is connected in series. Find R.
A. `700Omega`
B. `800Omega`
C. `900Omega`
D. `100Omega`

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1 Answers

Correct Answer - C
When a resistance of `100 Omega` is connected in series
current, `i=(V)/(100+R)` ….(i)
When a resistance of `100 Omega` is connected in series, the its range double
current, `i=(2V)/(1100+R)` …(ii)
From Eqs. (i) and (ii),
`(V)/(100+R)=(2V)/(1100+R)`
`rArr " " R = 900 Omega`

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