A galvanometer of resistance 20\(\Omega\) is to be provided with a shunt that \(\frac{1}{10}\) of the whole current in a circuit passes through the galvanometer. The resistance of the shunt is
The correct answer is: B
Explanation
\(\frac{1}{10}\)I x 20 = \(\frac{9}{10}\)I x RsRs = 2.22\(\Omega\)