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)=60.00 Omega, ` shunt resistance , `r_(s)=0.02 Omega`
total resistance in the circuit is `R_(G)+3=63 Omega`
Hence, `I=3//63 =0.048 A`
Resistance of the galvanometer converted to an ammeter is,
`(R_(G)r_(s))/(R_(G)+r_(s)) =(60Omegaxx0.02 Omega)/((60+0.02)Omega) =0.02 Omega`
total resistance in the circuit =0.02 +3=`3.02 Omega`
Hence, `I=3//3.02 =0.99 A`