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# A galvanometer together with an unknown resistance in series is connected 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 ........ ?

Akash Kumar Dutta
98 Points
8 years ago

Dear Yogita,

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 (ANS).

Regards.

Vikas TU
14149 Points
4 years ago
The net resistance = R+2r
Potential Difference(V) = 3V
As from the ohms law
•
3 = 1(R+2r)---------- 1.)
The net resistance in second case = R+r/2
V = 1.5V I =6A
1.5 = 6(R+r/2)
¼ =R+r/2 ---------(2.)
3r/2 = 11/4
r = 22/12
r =1.83 ohm