For the circuit (figure) the currents is to be measured. The ammeter shown is a galvanometer with a resistance `R_(G)=60.0 0 Omega` converted to an ammeter by a shunt resistance `r_(S)=0.02 Omega`. The value of the current is

Options

A. 0.79 A
B. 0.29 A
C. 0.99 A
D. 0.8 A

Answer

C

Explanation

`R_(G)=60 Omega`, shunt resistance, `r_(s)=0.2 Omega`
Resistance of the galvanometer converted to an ammeter is,
`(r_(G)r_(S))/(R_(G)+r_(s))=(60Omega xx 0.02 Omega)/((60+0.02)Omega)=0.02 Omega`
Total resistance in the circuit is given `= 0.02+3=3.02 Omega`
Hence, `l = 3//3.02 = 0.99 A`