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    				| 1. | 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? | 
| Answer» (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` | |