Question

An ammeter is obtained by shunting a $30\Omega $ galvanometer with a $30\Omega $ resistance. What additional shunt should be connected across it to double the range?
A) $15\Omega $
B) $10\Omega $
C) $5\Omega $
D) $\text{None of these}$

Answer 1

Hint: A type of resistor that has a very low value of resistance is known as shunt resistance. The shunt resistor is made up of a material that has a very low value of coefficient of resistance. The large value of current is measured by using a shunt resistor.

Complete step by step solution:
When the shunt resistor is connected in parallel to the ammeter then only a small amount of current flows through the ammeter. The rest of the current flows through the shunt resistor.
Given that the shunt resistance is $S = 30\Omega $
The reading of current in galvanometer is $G = 30\Omega $
The shunt is given by using the formula
$ \Rightarrow S = \dfrac{{{i_g}}}{{I - {i_g}}}G$
Where S is the shunt resistance
Substituting the values in the above equation and solving,
$30 = \dfrac{{{i_g}}}{{I - {i_g}}} \times 30$
$\dfrac{{{i_g}}}{{I - {i_g}}} = 1$
${i_g} = I - {i_g}$
$ \Rightarrow I = 2{i_g}$
When the range of the current is doubled then it can be written that
$ \Rightarrow I' = 2I$
$ \Rightarrow I = 2 \times 2{i_g}$
$ \Rightarrow I = 4{i_g}$
The required shunt will then become
$ \Rightarrow S' = \dfrac{{{i_g}}}{{I' - {i_g}}}G$
$ \Rightarrow S' = \dfrac{{{i_g}}}{{4{i_g} - {i_g}}} \times 30$
$ \Rightarrow S' = \dfrac{{{i_g}}}{{3{i_g}}} \times 30$
$ \Rightarrow S' = 10\Omega $
As shunt is connected in parallel to the galvanometer, if the additional shunt is x then
$ \Rightarrow \dfrac{{30x}}{{30 + x}} = 10$
$ \Rightarrow \dfrac{{30x}}{{10}} = 30 + x$
$\Rightarrow 3x = 30 + x$
$\Rightarrow 2x = 30$
$\Rightarrow x = 15\Omega $
The additional shunt connected with the ammeter to double the range is $ = 15\Omega $.

Option A is the right answer.

Note: It is important to note that the shunt is a device that is connected in parallel to the galvanometer. This is because the shunt does not allow a large amount of current to pass through the galvanometer and hence protects it from damage. When the set up of the circuit is such that the current is too high then the shunt resistor is connected in parallel.