In the circuit shown in figure, the current is to measured. What is the value of the current if the ammeter shown (i) is a galvanometer with a resistance `R_G=60*00Omega`. (ii) is a galvanometer described in (i) but converted to an ammeter by a shunt resistance `r_s=0*02Omega`, (iii) is an ideal ammeter with zero resistance?

Explanation

(a) Total resistance in the circuit is,
`R_(G) +3 = 63 Omega`. Hence, `I = 3//63 = 0.048A`.
(b) Resistance of the galvanometer converted to an ammeter is,
`(R_(G)r_(s))/(R_(G)+r_(s)) = (60 Omega xx 0.02 Omega)/((60+0.02)Omega) ~~ 0.02Omega`
Total resistance in the circuit is,
`0.02 Omega + 3 Omega = 3.02 Omeg`. Hence, `I = 3//3.02 = 0.99A`.
(c) For the ideal ammeter with zero resistance,
`I = 3//3 = 1.00A`