Question

When a galvanometer is shunted with a $4\Omega$ resistance, the deflection is reduced to $\dfrac{1}{5}$ . If the galvanometer is further shunted with a $2\Omega$ wire, the new deflection will be (assuming the main current remains the same):
A. $\dfrac{5}{{13}}$ Of the deflection when shunted with $4\Omega$ only
B. $\dfrac{8}{{13}}$ Of the deflection when shunted with $4\Omega$ only
C. $\dfrac{3}{4}$ Of the deflection when shunted with $4\Omega$ only
D. $\dfrac{3}{{13}}$ Of the deflection when shunted with $4\Omega$ only

Answer 1

In order to solve this question we need to understand the combination of resistance. So resistance is a physical device, which resists the current flow in circuit. So there are two combinations of resistances, one is series combination and other is parallel combination. In series combination, resistance is connected with each other in series such that current through resistances remain the same. In parallel combinations, resistance is connected with each other in parallel such that only current remains the same in both resistance.

Complete step by step answer:
Consider a galvanometer G having resistance ${R_g}$. Let us say that at this time deflection is $I$ so the current through the galvanometer is, $I$. Now suppose resistance $4\Omega$ (shunt resistance) is connected in parallel, let now current in the galvanometer is $\dfrac{I}{5}$ (According to problem).So the current in $4\Omega$ would be $\dfrac{{4I}}{5}$.

Since both galvanometer and resistance is connected in parallel so voltage across both would be same
So, ${V_{4\Omega }} = {V_{{R_g}}}$
$4\Omega \times \dfrac{{4I}}{5} = {R_g} \times \dfrac{{4I}}{5}$
${R_g} = 16\Omega$
Now consider the case when $2\Omega$ resistance is also connected in parallel with the galvanometer and $4\Omega$ resistance.
So the equivalent combination of $2\Omega$ and $4\Omega$ resistance is,
${R_e} = \dfrac{{2 \times 4}}{{2 + 4}}$
$\Rightarrow {R_e} = \dfrac{4}{3}\Omega$

Now the ${R_e}$ and galvanometer are connected in parallel, so current in arm containing galvanometer is given by,
${I_1} = \dfrac{{{R_e}}}{{{R_e} + {R_g}}}I$
Putting values we get,
${I_1} = \dfrac{{\dfrac{4}{3}}}{{16 + \dfrac{4}{3}}}I$
$\Rightarrow {I_1} = \dfrac{1}{{13}}I$
So the deflection in the galvanometer when additional $2\Omega$ resistance also connected in parallel with galvanometer and $4\Omega$ resistance is,
${I_1} = \dfrac{1}{{13}}I$
And deflection when only $4\Omega$ resistance in connected in parallel with galvanometer is, ${I_2} = \dfrac{1}{5}I$
Comparing the two we get,
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{(\dfrac{1}{{13}})I}}{{(\dfrac{1}{5})I}}$
$\therefore \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{5}{{13}}$

So the correct option is A.

Note: It should be remembered that current in parallel resistance is divided in inverse ratio of their resistance, and in series combination voltage is divided into resistance, in ratio of their resistance. So here we have used this concept to calculate the current in a galvanometer in case when both $2\Omega \& 4\Omega$ are connected in parallel.