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:
Answer & explanation
Correct answer: option 4
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]