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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.
How is that circuit wont be broken even there is an air gap in between the parallel plates? Ans) In AC circuit capacitor offers resistance to current as aresistor would do in DC.
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