Question

A galvanometer of resistance $40\;{\rm{\Omega }}$ and current passing through it is $100\;{\rm{\mu A}}$ per division. The full scale has 50 divisions. If it is converted into an ammeter of range 2 A by using a shunt, then the resistance of ammeter is
(1) $\dfrac{{40}}{{399}}\;{\rm{\Omega }}$
(2) $\dfrac{4}{{399}}\;{\rm{\Omega }}$
(3) $0.01\;{\rm{\Omega }}$
(4) $0.4\;{\rm{\Omega }}$

Answer 1

Hint: This question uses the concept of the galvanometer and the process to convert the galvanometer to the ammeter. First, you need to calculate the amount of current required to produce a full-scale reading in the galvanometer. Later, we need to substitute the values in the formula used for calculating the shunt. As you have only one unknown variable, you need to solve it.

Complete step by step answer:
Let us first look at the data given in the question. The resistance of the galvanometer is $G = 40\;{\rm{\Omega }}$ and the current passing through the galvanometer is $100\;{\rm{\mu A}}$ per division. The full scale of the galvanometer has 50 divisions. The amount of current passing through the ammeter ranges 2 A. This means that the galvanometer is converted into ammeter by using a shunt.
It is given that the galvanometer has 50 divisions. So the current required to produce full scale deflection can be calculated as,
$
{I_g} = 100\;{\rm{\mu A/division}} \times \dfrac{{{{10}^{ - 6}}\;{\rm{A}}}}{{1\;{\rm{\mu A}}}} \times\;{\rm{division}}\\
\Rightarrow {I_g} = 0.005\;{\rm{A}}
$
As we all know that the expression for the galvanometer being converted into ammeter is given as,
$\dfrac{I}{{{I_g}}} = 1 + \dfrac{G}{S}$
Here, I is the current through the ammeter, ${I_g}$ is the current through the galvanometer, G is the resistance of the galvanometer, and S is the resistance of the shunt.
As we know all the numeric values, we substitute the values in the above expression in order to get the values of resistance of the shunt (S).
$\dfrac{{2\;{\rm{A}}}}{{0.005\;{\rm{A}}}} = 1 + \dfrac{{40\;{\rm{\Omega }}}}{S}$
On further solving we get,
$\begin{array}{c}
400 = 1 + \dfrac{{40\;{\rm{\Omega }}}}{S}\\
\Rightarrow \dfrac{{40\;{\rm{\Omega }}}}{S} = 400 - 1\\
\Rightarrow S = \dfrac{{40}}{{399}}\;{\rm{\Omega }}
\end{array}$
Thus, the resistance of the ammeter is $\dfrac{{40}}{{399}}\;{\rm{\Omega }}$.

So, the correct answer is “Option A”.

Note:
Make sure that you calculate the current required to produce full-scale division and also make sure that you perform calculations after converting the units, or otherwise you will get the wrong answer.