Question

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

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

Answer 1

Answer: (C)

$R_{G}=60.00\, \Omega$, 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,
$\frac{R_{G} r_{s}}{R_{G}+r_{s}}=\frac{60\, \Omega \times 0.02\, \Omega}{(60+0.02) \Omega}=0.02 \,\Omega$
total resistance in the circuit $=0.02+3=3.02\, \Omega$
Hence, $I=3 / 3.02=0.99 \,A$