Question

A cell having an emf. of 2:2 V and internal resistance 0.2 ohm is connected to a circuit comprising an amneter and or sistance of 4 ohm in series with a combination of two resistances of 0.4 ohm each in parallel. What will be the reading of the am reter?​

Answer 1

Step-by-step explanation:

e.m.f=I(R+r)

e.m.f=2.2V

I=I

r=0.2ohm

R= R1+R2

R1=4 ohm

R2=1/0.4+1/0.4=1/5ohm

I=e.m.f/(R+r)

I=2.2(0.2+4+1/5)

I=2.2(2.4)

I=5.28A