Question

A galvanometer of resistance $20\mathrm{\Omega }$ is to be shunted so that only $1\%$ of the current passes through it. Shunt connected is

• A. $\frac{99}{20}\mathrm{\Omega }$
• B. $\frac{9}{20}\mathrm{\Omega }$
• C. $\frac{20}{99}\mathrm{\Omega }$
• D. $\frac{2}{99}\mathrm{\Omega }$

Answer 1

The correct option is C

$\frac{20}{99}\mathrm{\Omega }$

Step 1. Given data

A galvanometer of resistance $20\mathrm{\Omega }$ is to be shunted so that only $1\%$ of the current passes through it.

We have to calculate the required shunt.

Step 2. Formula to be used

A galvanometer works on the principle of electromagnetic induction, which is when a current-carrying conductor is placed in a magnetic field, it experiences a torque.

The general formula to calculate the shunt is,

$\mathrm{S}=\frac{\mathrm{G}}{\left({\displaystyle \frac{\mathrm{i}}{{\mathrm{i}}_{\mathrm{g}}}}-1\right)}$

Here, ${\mathrm{i}}_{\mathrm{g}}$ is the current passing through the galvanometer, $\mathrm{S}$ is shunt, $\mathrm{G}$ is galvanometer constant, and $\mathrm{i}$ is the total current in the circuit

Step 3. Find the value.

Let shunt connected be $\mathrm{R}\mathrm{\Omega }$ in parallel,

${\mathrm{i}}_{\mathrm{g}}=1\%$ of $i$

$=\frac{1}{100}i$

Now, we will calculate the required shunt by using the formula of galvanometer into an ammeter.

So,

$\mathrm{S}=\frac{\mathrm{G}}{\left({\displaystyle \frac{\mathrm{i}}{{\mathrm{i}}_{\mathrm{g}}}}-1\right)}$

Here,

$\mathrm{G}=20\mathrm{\Omega }$

${\mathrm{i}}_{\mathrm{g}}=\frac{\mathrm{i}}{100}$

Substitute all the values in the above formula, we get,

$$\begin{gathered} \mathrm{S}=\frac{20}{\left({\displaystyle \frac{i}{\left({\displaystyle \frac{i}{100}}\right)}}-1\right)} \\ \Rightarrow \mathrm{S}=\frac{20}{\left({\displaystyle 100-1}\right)} \\ \Rightarrow \mathrm{S}=\frac{20}{99}\Omega \end{gathered}$$

Shunt connected is $\frac{20}{99}\Omega$

Hence, option C is the correct answer.