Question

In the adjoining circuit the battery E1 has an emf of 12 volt and zero internal resistance.while the battery E has an emf of 2 volt .if galvanometer reads zero then the value of the restiance X in tga ohm​

Answer 1

Answer:

Consider the circuit as shown in figure.

When there is no current in galvanometer, the potential difference between PQ will be same as between BC.

Hence, V

PQ

=2 V

Also the current through the loop APQDA is

I=

500+x

12

According to Ohm's law,

V

PQ

=Ix

2=(

500+x

12

)x

6x=500+x

∴x=100Ω