Question

A moving coil galvanometer has a resistance of $900\Omega$. In order to send only 10% of the main current through galvanometer, the resistance of the required shunt is:
A. $0.9\Omega$
B. $100\Omega$
C. $405\Omega$
D. $90\Omega$

Answer 1

Shunt resistance is a resistor with such a type of resistor having a very low resistance value is called shunt resistance. The shunt resistor is usually constructed of a substance with a resistance coefficient of low temperature. It is related to the ammeter, whose range is to be expanded, in parallel.
Formula used:
For solving this question, we will be using the formula for the Shunt resistance, i.e.,
$S=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$

Complete answer:
Let us take a look at all the given parameters,
G=$900\Omega$
Let the main current be I
Now, current through galvanometer will be 10% of the main current,
$\Rightarrow {{I}_{g}}=0.1I$
So, current through shunt will be 90% of the main current,
$\Rightarrow {{I}_{S}}=0.9I$
Now, applying the formula for the resistance of the shunt, that we discussed above
$S=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$
Using the given parameters in the formula for the resistance of shunt
$\Rightarrow S=\dfrac{0.1I\times 900}{I-0.1I}$
$\Rightarrow S=100\Omega$
So, in order to send only 10% of the main current through the galvanometer.

The resistance of the required shunt will be Option – B, i.e., $100\Omega$.

Note:
The galvanometer is an electromechanical device used to detect and detect electrical current. The galvanometer operates as an actuator, producing a rotary deflection ("pointer") in response to the electrical current flowing through the coil in a constant magnetic field. The first galvanometer was reported by Johann Schweigger at the University of Halle on 16 September 1820. André-Marie Ampère has also contributed to its growth. Early designs improved the influence of the magnetic field produced by the current by using several wire turns.