Question

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

Answer 1

Hint: Given that some percentage of main current passes through the galvanometer, this is a case of shunt resistance. As a result, the circuit will have a low resistance value.

Formula Used: The given circuit arrangement is in parallel. Therefore, the concept of parallel connection considering, same potential difference and different current will be used. Hence according to Ohm’s law,
\[V = IR\]
Where V is the potential difference which is the same for parallel combination of resistors, I is the current and R is the resistance.
Also, to find out the equivalent resistance across a parallel circuit, the formula used will be
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ......\]

Complete step by step solution:
Given that the resistance of the galvanometer is G. Also current through the galvanometer is \[0.2\% \] of main current. That is,
\[{I_G} = \dfrac{{0.2}}{{100}}I = \dfrac{1}{{500}}I\]
The current through the shunt will be
\[{I_S} = I - \dfrac{1}{{500}}I\]
\[\Rightarrow {I_S} = \dfrac{{500I - I}}{{500}}\]
\[\Rightarrow {I_S} = \dfrac{{499}}{{500}}I\]

Since, shunt and galvanometer are connected in parallel, therefore, potential difference will be same for both. It can be written that
\[{V_S} = {V_G}\]
Using Ohm’s Law,
\[{I_S}{R_S} = {I_V}{R_V}\]
Substituting the values, we get
\[\dfrac{{499}}{{500}}I \times S = \dfrac{1}{{500}}I \times G\]
\[\Rightarrow S = \dfrac{G}{{499}}\]

Since, the circuit is in parallel connection, therefore, the equivalent resistance will be the reciprocal sum of individual resistances. The resistance across ammeter (shunt) is given by
\[\dfrac{1}{{{R_A}}} = \dfrac{1}{{{R_G}}} + \dfrac{1}{{{R_S}}}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{1}{G} + \dfrac{1}{S}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{1}{G} + \dfrac{{\dfrac{1}{G}}}{{499}}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{{500}}{G}\]
\[\therefore {R_A} = \dfrac{G}{{500}}\]

The resistance of the ammeter will be \[\dfrac{G}{{500}}\].

Note: It is important to remember that shunt is a device which, when there is a connection between two points, provides a low resistance path to current to flow through. It is used to measure current through the circuit. Since it creates a low resistance path for current in the circuit, therefore, it is also used to protect the galvanometer from high current.