Question

A galvanometer having a resistance of $9$ ohm is shunted by a wire of resistance $2$ ohm. If the total current is $1$ amp, the part of it passing through the shunt will be

• A. 0.2 amp
• B. 0.8 amp
• C. 0.25 amp
• D. 0.5 amp

Answer 1

Answer: (B)

The shunt and galvanometer are in parallel.
Therefore, $\frac{1}{R_{e q}}=\frac{1}{9}+\frac{1}{2}$
or, $R_{e q}=\frac{18}{11} \Omega$
Using Ohm's law, $V=I R_{e q}=I \times \frac{18}{11}=\frac{18}{11} V$.
$\therefore$ Current through shunt $=\frac{V}{R_{s}}$
$\frac{18 / 11}{2}=\frac{9}{11} \simeq 0.8 a m p$