Question

When a galvanometer is shunted with a $4\Omega$ resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with is further shunted with a $2\Omega$ wire $(2\Omega and4\Omega$ resistors are connected in parallel), the further reduction in the deflection will be(the main current remains the same).

• A. $\left(\frac{8}{13}\right)$ of the deflection when shunted with $4\Omega$ only
• B. $\left(\frac{5}{13}\right)$ of the deflection when shunted with $4\Omega$ only
• C. $\left(\frac{3}{4}\right)$ of the deflection when shunted with $4\Omega$ only
• D. $\left(\frac{3}{13}\right)$ of the deflection when shunted with $4\Omega$ only

Answer 1

The correct option is A

$\left(\frac{8}{13}\right)$ of the deflection when shunted with $4\Omega$ only

Case I : $R_g\times \frac{1}{5}=(I-\frac{1}{5})\times 4\Rightarrow R_g=16\Omega$
Case II : $16l_1=\frac{4\times 2}{6}(l-l_1)\Rightarrow l_1=\frac{l}{13}$
So decrease in current to previous current $=\frac{\frac{l}{5}-\frac{l}{13}}{\frac{l}{5}}$