A galvanometer has a resistance of 98 Ω. If 2% of the main current is to be passed through the meter, what should be the value of the shunt ?

Q: A galvanometer has a resistance of 98 Ω. If 2% of the main current is to be passed through the meter, what should be the value of the shunt?

Sol: G = 98 Ω ; $\large \frac{i_g}{i} \times 100 = 2 \% $

$\large i_g G = i_s S $

$\large S = \frac{i_g G}{i_s} = \frac{i_g G}{i – i_g} $

$\large S = \frac{G}{i/i_g – 1}= \frac{98}{50-1} $

S = 2 ohm