The maximum current that can be measured by a galvanometer of resistance 40 ~ , is 10 ~mA . It is converted into a voltmeter that can read upto 50 ~V . The resistance to be connected in series with the galvanometer (in ) is

Step-by-step solution

Given, G=40 , I_ g =10 ~mA =10 10⁻³ ~A V=50 ~V Let R be the resistance connected in series with the galvanometer to convert into voltmeter, then aligned & & V &=I_ g (G+R) & & 50 &=10 10⁻³(40+R) & & 50 10 10⁻³ &=40+R & & R &=5000-40=4960 aligned