Question

1) in the circuit the P.D. across AB is 3 volt and if internal impedance of the cell is 1 ohm law then it's EMF of the battery is ?​

Answer 1

Answer:

Correct option is

C) 10V

Let current in circuit =i

Then

PDb/wAB=iR=3V

                      =i×6=3

                    i=63=21A

In loop ABCDA

−6i−6i−7i−i−E=0

E=20i

E=(20×21)v

E=10v.

HEY HERE IS YOUR ANSWER!!!!