Question

It is required to convert a galvanometer of current range of 15mA and voltage range 750mV into an ammeter of range 25A. The value of necessary resistance that must be connected in parallel with the galvanometer is?
(A). ${\text{0}}{\text{.4}}\Omega$
(B). ${\text{0}}{\text{.03}}\Omega$
(C). $1\Omega$
(D). ${\text{0}}{\text{.5}}\Omega$

Answer 1

Hint: A galvanometer is an electrical device which is used to measure the electric current and voltage in a circuit . It is an electromechanical device in which there is a coil which gets energized when current flows through it and the needle attached to it gets deflected according to the magnitude of the current flowing through it and shows the correct reading in the dial gauge. To convert this galvanometer into an ammeter we have to connect a shunt resistance in parallel with the galvanometer then this combination will act as an ammeter.

Complete step-by-step solution -
Given,
Maximum current for galvanometer =${{\text{I}}_g}$=15mA=0.015A
Maximum voltage for galvanometer=${{\text{V}}_g}$=750mV=0.75V
Maximum current through the ammeter which is to be made with the help of galvanometer by adding the resistance in parallel=${\text{I}}$=25A,
To find-
The resistance which can be connected in parallel with the galvanometer=$S$=?
Firstly we will find the resistance of the galvanometer as follows,
Resistance of the galvanometer=
${\text{G}}$= $\dfrac{{{{\text{V}}_g}}}{{{{\text{I}}_g}}}$
=$\dfrac{{.75}}{{.015}}$
${\text{G}}$ =$50\Omega$
Now we will use the following relation,
$\dfrac{{{{\text{I}}_g}}}{{\text{I}}} = \dfrac{S}{{S + G}}$
$\Rightarrow$ $\dfrac{{.015}}{{25}} = \dfrac{S}{{S + 50}}$
$\Rightarrow$ $\dfrac{{25}}{{.015}} = \dfrac{{S + 50}}{S}$
$\Rightarrow$ $1666.667 = 1 + \dfrac{{50}}{{\text{S}}}$
$\Rightarrow$ $1665.667 = \dfrac{{50}}{S}$
$\Rightarrow$ ${\text{S = }}\dfrac{{50}}{{1665.667}}$
$\Rightarrow$ $S = 0.03001\Omega$
$\Rightarrow$ $S = 0.03\Omega$
Hence the value of the required resistance which is to be connected in parallel with the galvanometer is $0.03\Omega$.
So the option (B) is the correct answer.

Note: Galvanometer is basically used to detect the presence of the small current or voltage. It can also measure the magnitude of the current and voltage flowing through the circuit. Galvanometer can show the direction and magnitude of both of the current whereas the ammeter only shows the magnitude of the current in the circuit. Limitation regarding the galvanometer is that it can only measure the small current so to measure the large current we need ammeter. We can convert the galvanometer into ammeter or voltmeter. For converting the galvanometer into ammeter we connect the a shunt resistance of small magnitude in parallel with the galvanometer. If we want to change the galvanometer into the voltmeter then we will connect a large resistance known as multiplier in series with the galvanometer.