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.

Options

A. `900 Omega`
B. `100 Omega`
C. `500 Omega`
D. `800 Omega`

Answer

D

Explanation

Let I be the maximum current in the galvanometer of resistance G then `i=V_1/R_1=V_2/R_2 where V_1=V,V_2=2V,R_1=G+100,R_2+G+1000 thereforeV/(G+100)=(2V)/(G+1000)therefore2(G+100)=G+1000 therefore 2G-G=1000-200 therefore G=800Omega`.