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When a resistance of 100`Omega` is connected in series with a galvanometer of resistance R, then its range is V. To double its range, a resistance of 1000`Omega` is connected in series. Find the value of R.A. `700 Omega`B. `800 Omega`C. `900 Omega`D. `100 Omega` |
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Answer» Correct Answer - C When a resistance of `100 Omega` is connected in series current , i= `(V)/(100 + R) " " …. (i)` when a resistance of `1000 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)` `R = 900 Omega` |
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