Question

A galvanometer having a resistance of $50\Omega $ gives a full scale deflection for a current of \[0.05A\] . The length in meter of a resistance wire of area of cross section $2.97 \times {10^{ - 3}}c{m^{ - 2}}$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5A$ current is:
[specific resistance of wire$ = 5 \times {10^{ - 7}}\Omega m$ ]
A) $9$
B) $6$
C) $3$
D) $1.5$

Answer 1

Hint: In the given question, we need to convert the galvanometer into an ammeter. We know that an ammeter must have a very small resistance. In order to convert a galvanometer into an ammeter, we need to connect a very small shunt resistance in parallel to the galvanometer resistance to make the effective resistance value less. So, we will calculate the shunt resistance by using the formula which is $S = \dfrac{{{I_g}G}}{{I - {I_g}}}$ . After that we can calculate the length of the resistance wire by using the resistance formula which is $S = \dfrac{{\rho l}}{A}$ . By putting all the given values, we can calculate the required unknown parameter.

Formula used:
$S = \dfrac{{{I_g}G}}{{I - {I_g}}}$
Where,
$S = $ Shunt resistance
\[{I_g} = \] Galvanometer current
$G = $ Galvanometer resistance
$I = $ Maximum current that an ammeter can read

$S = \dfrac{{\rho l}}{A}$
Where,
$S = $ Shunt resistance
\[\rho = \] Specific resistance of wire
$l = $ Length of resistance wire
$A = $ Area of resistance wire

Complete step by step solution:
According to the question, we are given with the galvanometer resistance and current.

In order to convert the galvanometer into an ammeter, we need to find the shunt resistance which is to be connected in the parallel with the galvanometer resistance.
The shunt resistance can be calculated by:
$S = \dfrac{{{I_g}G}}{{I - {I_g}}}$
Now, putting the values in the above formula.
$
   \Rightarrow S = \dfrac{{{I_g}G}}{{I - {I_g}}} \\
   \Rightarrow S = \dfrac{{0.05 \times 50}}{{5 - 0.05}} \\
   \Rightarrow S = \dfrac{{2.5}}{{4.95}} \\
   \Rightarrow S = \dfrac{{250}}{{495}} \\
   \Rightarrow S = \dfrac{{50}}{{99}} \\
 $
As we have the shunt resistance, we can calculate the length of the wire by using a resistance formula.
$S = \dfrac{{\rho l}}{A}$
Now, putting the values given in the question.
\[
   \Rightarrow S = \dfrac{{\rho l}}{A} \\
   \Rightarrow l = \dfrac{{SA}}{\rho } \\
   \Rightarrow l = \dfrac{{50}}{{99}} \times \dfrac{{3 \times {{10}^{ - 6}}}}{{5 \times {{10}^{ - 7}}}} = 3.0m \\
 \]
Therefore, the length of the resistance wire is $3m$ .

Hence, the correct option is (C).

Note: In the given question, we are provided with the galvanometer parameters. When we need to convert the galvanometer into an ammeter, we have to minimise the resistance which can be done by adding a shunt resistance in parallel. After minimizing the resistance, we can calculate the length of wire as we have the area of cross section and the specific resistance of wire. We have to be careful while putting the values in the formulas.