Question

A permanent magnet moving coil gives full scale deflection at $40\,{\text{mV}}$ potential difference and $8\,{\text{mA}}$ current. What will be the required series resistance when it is used as a voltmeter of range $0 - 200\,{\text{V}}$ ?
(A) $19556\,{\text{ohm}}$
(B) $20163\,{\text{ohm}}$
(C) $23884\,{\text{ohm}}$
(D) $24995\,{\text{ohm}}$

Answer 1

First of all, we will find the resistance of the galvanometer. Then we will find the resistance of the shunt resistor by substituting the required values and manipulating accordingly.

Complete step by step answer:
In the given question, the data provided are as follows:
A permanent magnet moving coil gives full scale deflection at a potential difference $40\,{\text{mV}}$ .
The current at which it gives full scale deflection is $8\,{\text{mA}}$ .
The magnet moving coil when used as a voltmeter has a range of $0 - 200\,{\text{V}}$ .
For better understanding, we draw a circuit diagram, where the galvanometer and the shunt are connected in series and the overall connection is connected to the voltmeter in parallel.
As the device gives full scale deflection at potential $40\,{\text{mV}}$ and electric current $8\,{\text{mA}}$ .
So, the resistance of the galvanometer will be:
By Ohm’s law, we have:
${R_{\text{g}}} = \dfrac{V}{I}$ …… (1)
Where,
${R_{\text{g}}}$ indicates the resistance of the galvanometer.
$V$ indicates potential difference.
$I$ indicates electric current.
Substituting, the required values in the equation (1), we get:

$${R_{\text{g}}} = \dfrac{{40\,{\text{mV}}}}{{8\,{\text{mA}}}} \\\ {R_{\text{g}}} = 5\,\Omega \\\$$

The resistance of the galvanometer is found to be $5\,\Omega$ .
Now, we need to find the resistance of the shunt so that the potential difference across is equal to $200\,{\text{V}}$ , then the potential difference across the galvanometer is $40\,{\text{mV}}$ to give the full deflection.
Now, we apply the formula:
${V_{\text{g}}} = \dfrac{{{R_{\text{g}}}}}{{S + {R_{\text{g}}}}}{\text{V}}$ ……….(2)
Where,
${V_{\text{g}}}$ indicates potential difference of the galvanometer.
${R_{\text{g}}}$ indicates the resistance of the galvanometer.
$S$ indicates the resistance of the shunt.
${\text{V}}$ indicates the potential difference of the voltmeter.
Substituting the required values in the equation (2) and we get:

$${V_{\text{g}}} = \dfrac{{{R_{\text{g}}}}}{{S + {R_{\text{g}}}}}{\text{V}} \\\ {\text{40}} \times {\text{1}}{{\text{0}}^{ - 3}} = \dfrac{5}{{S + 5}} \times 200 \\\ S + 5 = \dfrac{{1000}}{{{\text{40}} \times {\text{1}}{{\text{0}}^{ - 3}}}} \\\ S = 25000 - 5 \\\$$

$S = 24995\,\Omega$
The required value of the resistance is $24995\,\Omega$ .
The correct option is D.

Note: It is important to remember that the shunt resistance is always connected to the galvanometer. Galvanometer is used in the circuit (in series) to detect a small current. Shunt resistors are used to measure current.