In the circuit the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance `R_G = 60.00 Omega`, (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance `r_s= 0.02 Omega `, (c ) is an ideal ammeter with zero resistance?

Explanation

(a)
For a single closed loop,
Current = `("Effective emf")/("Equivalent resistance")`
`thereforeI=epsi/(R+G)" "...(1)`
`thereforeI=3/(3+60)=3/63=0.0476A`
(b)
Here, resistance of ammeter will be,
`R_(A)=(SG)/(S+G)=(SG)/(G(S/G+1))=S/(S/G+1)" "...(2)`
here, `SltltltltGrArrS/Gltltltlt1`
Hence, `R_(A)~~S=0.02Omega`
[By neglecting `S/G` in equation (2)]
Now, above circuit reduces to single closed loop as follows.

`I.=epsi/(R+R_(A))" "...(3)`
`thereforeI.=3/(3+0.02)`
`thereforeI.=3/3.02=0.9934A`
( c) Since ideal ammeter has no resistance at all substituting `R_(A)=0` in equation (3), we get
`I..=epsi/R=3/3=1A`