A galvanometer having a resistance of 8 Ω is shunted by a wire of resistance 2 Ω. If the current is 1 A, the part of it passing through the shunt will be:

0.8 A

The correct answer is Option (4) → 0.8 A

The total resistance $R_{total}$ for two resistor in parallel -

$\frac{1}{R_{total}}=\frac{1}{R_G}+\frac{1}{R_S}$

$=\frac{1}{8}+\frac{1}{2}=\frac{5}{8}$

$⇒R_{total}=\frac{8}{5}=1.6Ω$

Now,

Current passing through shunt,

$I_s=I×\frac{R_g}{R_g×R_s}=1×\frac{8}{8+2}=0.8A$ [By Ohm's law]