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, $\frac{1}{R_{eq}}=\frac{1}{9}+\frac{1}{2}$
or, $R_{eq}=\frac{18}{11}\Omega$
Using Ohm's law, $V=I{R}_{eq}=I\times \frac{18}{11}=\frac{18}{11}V$.
$\therefore$ Current through shunt $=\frac{V}{R_s}$
$\frac{18/11}{2}=\frac{9}{11}\simeq 0.8amp$