In the circuit shown below, E is a battery of emf 6 V and internal resistance `1Omega`. Find the reading of the ammeter A. if it has negligible resistance :

Explanation

In the given circuit diagram, `R_(2),R_(6)` and `R_(4)` are in series. So,
`R_(1).=7+5+12=24Omega`
Now `R_(1).` and `R_(5)` are in parallel. So,
`(1)/(R_(2).)=(1)/(8)+(1)/(24)=(3+1)/(24)`
`=(4)/(24)=(1)/(6)`
`R_(2).=6Omega`
Now, `R_(2).,R_(1)` and `R_(3)` are in series. So,
`R=R_(2).+R_(1)+R_(3)`
`=6+3+2=11Omega`
We know, `i=(E)/(R+r)`
`(6)/(11+1)`
`=(6)/(12)=(1)/(2)`
`i=0.5A`