Question

A galvanometer having a total resistance of 8 ohm is shunted by a wire of resistance 2 ohm to convert it in to an ammeter. If the total current in to the meter is I, then the part of the current $I_s$ flowing through the shunt is

• A. $I_s$ $=$ 0.2 I
• B. $I_s$ $=$0.8 I
• C. $I_s$ $=$2 I
• D. $I_s$ $=$8I

Answer 1

Answer: (B)

$I_s$ $=$0.8 I

$Req=\frac{8\times 2}{8+2}=1.6\Omega$
$\left(1.6\right)I=I_s\left(2\right)$
= $I_s=0.8I$