For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance RG=60.00 Ω converted to an ammeter by a shunt resistance rs=0.02 Ω. The value of the current is

Question

For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance RG=60.00 Ω converted to an ammeter by a shunt resistance rs=0.02 Ω. The value of the current is (A) 0.79 A (B) 0.29 A (C) 0.99 A (D) 0.8 A

Tardigrade Admin · Accepted Answer

RG=60.00 Ω, shunt resistance, rs=0.02 Ω total resistance in the circuit is RG+3=63 Ω Hence, I=3 / 63=0.048 A Resistance of the galvanometer converted to an ammeter is, (RG rs/RG+rs)=(60 Ω × 0.02 Ω/(60+0.02) Ω)=0.02 Ω total resistance in the circuit =0.02+3=3.02 Ω Hence, I=3 / 3.02=0.99 A

Q. For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance RG​=60.00Ω converted to an ammeter by a shunt resistance rs​=0.02Ω. The value of the current is

0.79 A

0.29 A

0.99 A

0.8 A

Q. For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance $R_{G} = 60.00 Ω$ converted to an ammeter by a shunt resistance $r_{s} = 0.02 Ω$ . The value of the current is