In the circuit (Figure ) , the current it to be measured. What is the value of the current if the ammeter shown (a) is galvanometer with a resitance `R_(G) = 60,00 Omega`, (b) is a galvanometer by a shunt resitance `r_(s) = 0.02 omega` (c) is an ideal ammter with zero resitance ?

Explanation

(a) Total resistance in the circuit is
`R_G + 3 = 63 Omega`. Hence `I = 3/63 = 0.048 A`
(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) ~~ 0.02 Omega`
Total resistance in the circuit is
`0.02 Omega + 3 Omega = 3.02 Omega. ` Hence I = 3/3.02 = 0.99 A
(c ) For the ideal ammeter with zero resistance .
I = 3/3 = 1.00 A