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A galvanometer together with an unknown resistance in series is linked across two identical batteries each of 1.5V. whenthese batteries are connected in series the galvanometer records a current of 1A and when the batteries are in parallel the current is 0.6A. The internal resistance of battery will be ?
Ans) R(net) in first case= R+2r
V=3V,
I=1=V/R=>R+2r=3.......eq(1)
R(net) in Second case= R + r/2
V=1.5V
I=V/R=>.6=3/2R+r
=> 5=2R+r..............eq(2)
Solving eq1 and eq2
we get r= 1/3 or .33 ohms.
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