Question

A galvanometer with resistance $100\Omega$ gives full scale deflection with a current of 10 mA. The value of shunt, in order to convert it into an ammeter of 10 Ampere range, will be
$A. -10\Omega$
$B. 1 \Omega$
$C. 0.1\Omega$
$D. 0.01 \Omega$

Answer 1

To solve this question, use the formula for total current flowing through the galvanometer circuit which is equal to the sum of current through galvanometer and current through shunt resistance. Rearrange this equation and obtain an equation for current through shunt resistance. Now, to convert a galvanometer into an ammeter, potential difference across the galvanometer should be equal to the potential difference across the shunt resistance. Obtain an expression for this, substitute the values in that expression and find the value of shunt resistance.

Formula used:
$I= {I}_{g}+ {I}_{s}$

Complete answer:
Given: $G= 100\Omega$
$I = 10 A$
${I}_{g}= 10 mA= 0.01 A$
The total current flowing through a galvanometer circuit is given by,
$I= {I}_{g}+ {I}_{s}$ …(1)
Where, ${I}_{g}$ is the current flowing through the galvanometer
${I}_{s}$ is the current flowing through the shunt resistance
Rearranging equation. (1) we get,
${I}_{s}= I- {I}_{g}$ …(2)
To convert into an ammeter, a shunt resistance and galvanometer has to be connected in parallel. Thus, potential difference across the galvanometer= potential difference across the shunt resistance.
$\therefore {I}_{g} \times G= {I}_{s} \times S$ …(3)
Where, S is the shunt resistance
G is the galvanometer resistance
Substituting equation. (2) in equation. (3) we get,
${I}_{g} \times G= \left(I- {I}_{g}\right) \times S$
Substituting values in above equation we get,
$0.01 \times 100= \left(10-0.01\right) \times S$
$\Rightarrow 1= 9.99 \times S$
$\Rightarrow S= \dfrac {1}{9.99}$
$\Rightarrow S= 0.1 \Omega$
Thus, the value of shunt will be $0.1 \Omega$.

So, the correct answer is option C i.e. $0.1 \Omega$.

Note:
Students must remember how to convert a galvanometer into an ammeter or a voltmeter. To convert a galvanometer into an ammeter, a very small shunt resistance has to be connected in parallel always. While to convert a galvanometer into a voltmeter, a large resistance has to be connected in series. The galvanometer resistance is considered as a simple resistance which is in series with an ideal galvanometer. An ideal galvanometer has zero resistance.