Question

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

Answer 1

Hint : In an ammeter, a galvanometer and a shunt resistance are connected in parallel. The resistance of the ammeter will be equal to the net parallel resistance of the galvanometer and the shunt resistance.

Formula used: In this solution, we will use the following formula
$\Rightarrow {R_{net}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$ where ${R_{net}}$ is the net resistance of two resistors ${R_1}$ and ${R_2}$ connected in parallel.

Complete step by step answer
We’ve been given that an ammeter $0.2\%$ of main current passes through the galvanometer. So, if the total current is $I$ , the current through the of the galvanometer ammeter will be:
$\Rightarrow {I_G} = \dfrac{{0.2}}{{100}}I$
Since the current branches in a parallel connection, the current in the shunt is:
$\Rightarrow {I_S} = \dfrac{{99.8}}{{100}}I$
Now, for a parallel connection, we know that the voltage drop across the two parallel branches will be the same i.e.
$\Rightarrow {V_S} = {V_G}$
So, from ohm’s law we can write:
$\Rightarrow {I_S}{R_S} = {I_G}{R_G}$
Since the resistance of the galvanometer is $G$ that is ${R_G} = G$ , we can write
$\Rightarrow {I_S}{R_S} = {I_G}G$
Substituting the value of ${I_G} = \dfrac{{0.2}}{{100}}I$ and ${I_S} = \dfrac{{99.8}}{{100}}I$ , we get
$\Rightarrow \dfrac{{99.8}}{{100}}I{R_S} = \dfrac{{0.2}}{{100}}IG$ $$
Cancelling out $I$ from both sides and solving for ${R_S}$ , we get
$\Rightarrow {R_S} = G/499$
The net resistance of the ammeter will be
$\Rightarrow {R_A} = \dfrac{{{R_S}{R_G}}}{{{R_S} + {R_G}}}$
Substituting ${R_S} = 499G$ and ${R_G} = G$ , we get
$\Rightarrow {R_A} = \dfrac{{\dfrac{G}{{499}} \times G}}{{\dfrac{G}{{499}} + G}}$
This gives us
$\Rightarrow {R_A} = \dfrac{{{G^2}}}{{499}} \times \dfrac{{499}}{{500G}}$
Cancelling similar terms we get,
$\therefore {R_A} = \,G/500$
Hence the resistance of the ammeter will be $G/500$ .

Note
We must be aware of the internal construction of an ammeter which has a galvanometer and a shunt resistor in it connected in parallel to each other. Then we can easily solve the current flowing and the resistance of different components using the connection of series and parallel connection of resistors.