Question

In an ammeter 0.2% of main current passes through the galvanometer. If the resistance galvanometer is G, the resistance of the ammeter will be.

Answer 1

Hint: In this question first we draw an ammeter by connecting a resistance ${R_{sh}}$ in a shunt across the galvanometer whose resistance is given as $G\Omega $. Then equating the potential as they are connected in parallel that is ${V_{Rsh}} = {V_G}$ we get the value of shunt resistance as ${R_{sh}} = \dfrac{2}{{998}} \times G$. Then we solve for total ammeter resistance that is ${R_A} = {R_{sh}}\parallel G$ and get the required answer.

Complete Step-by-Step solution:
The first step is to know how an ammeter is constructed using a galvanometer. When we connect a parallel resistance across the galvanometer with very low resistance. And as mentioned in the question that only 0.2% of the main current is passing through the galvanometer. So we have drawn the ammeter with all the current components as shown in figure 1.

Figure1

Here, let us assume the main current be \[I\]
So current through galvanometer is ${I_G} = 0.2\% I = \dfrac{2}{{1000}}I$
Now the current through the shunt resistance ${R_{sh}}$ is
 ${I_{sh}} = I - {I_G} = I - \dfrac{2}{{1000}}I$
$ \Rightarrow {I_{sh}} = \dfrac{{998}}{{1000}}I$
Now we are given that the resistance of the galvanometer is $G\Omega $, So to find the overall resistance of ammeter ${R_A}$ that is the parallel combination of ${R_{sh}}$ and $G\Omega $, we need to calculate ${R_{sh}}$.
${R_A} = {R_{sh}}\parallel G$
From the figure we can see that the potential across ${R_{sh}}$ is same as potential across $G$ that is
${V_{Rsh}} = {V_G}$
$ \Rightarrow {I_{sh}} \times {R_{sh}} = {I_G} \times G$
$ \Rightarrow \dfrac{{998}}{{1000}}{I} \times {R_{sh}} = \dfrac{2}{{1000}}{I} \times G$
$ \Rightarrow {R_{sh}} = \dfrac{2}{{998}} \times G$
Now ${R_A} = {R_{sh}}\parallel G$
\[ \Rightarrow {R_A} = \dfrac{2}{{998}}G\parallel G\]
$ \Rightarrow {R_A} = \dfrac{1}{{499}}G\parallel G$
Now by solving this we will get
$ \Rightarrow {R_A} = \dfrac{{\dfrac{1}{{499}}G \times G}}{{\dfrac{1}{{499}}G + G}} = \dfrac{{{{{G}}^2}}}{{{G}\left( {499 + 1} \right)}}$
$ \Rightarrow {R_A} = \dfrac{G}{{500}}\Omega $
Therefore the overall ammeter resistance is $\dfrac{G}{{500}}\Omega $

Note: For these kinds of questions we first need to know how to construct an ammeter, voltmeter, and wattmeter using a galvanometer. Then we can find the overall resistances either by parallel connection or by the series connection of all the resistance that is present. We also need to have a good knowledge of KVL and KCL.