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Grade 12th passElectric Current

Galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1.5 holes in the batteries are connected in series the Galvanometer records a current of 1 amps and when the batteries are connected in parallel the current is 6 amps what is internal resistance of each battery

Janardhan P

9 Years agoGrade 12th pass

1 Answer

Vikas TU

9 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

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