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Question: A galvanometer having a coil of \(12\Omega \) gives full scale deflection for a current of \(4\)mA. ...

A galvanometer having a coil of 12Ω12\Omega gives full scale deflection for a current of 44mA. How can it be converted into a voltmeter of range 0 to 24 V?

Explanation

Solution

A galvanometer is an electromagnetic device that can be used to measure currents even up to the order of a few microamperes. This galvanometer device is used to detect the presence of current or measure its magnitude in a circuit. To solve this question use the equation of voltage drop across a branch consisting of a series combination of the galvanometer and a high resistance. Then by substituting the value of galvanometer resistance, the current through the coil and the voltmeter value we can determine the value of the resistance.

Complete step-by-step solution:
The given values are:
Resistance of galvanometer RGR_G =12Ω= 12\Omega
Current of galvanometer IG_G=44mA= 41034*{10^{ - 3}}A
Voltage= 2424 V
To find: Resistance R
Using the equation: V= IG(R+RG)V = {\text{ }}{I_G}\left( {R + {R_G}} \right)
Rearranging we get, VIG=(R+RG)\dfrac{V}{{{I_G}}} = \left( {R + {R_G}} \right)
R=VIGRGR = \dfrac{V}{{{I_G}}} - {R_G}
Substituting the values of resistance, voltage and current we get, R=24410312R = \dfrac{{24}}{{4*{{10}^{ - 3}}}} - 12
Simplifying the values we get, R=610312=600012=5988ΩR = \dfrac{6}{{{{10}^{ - 3}}}} - 12 = 6000 - 12 = 5988\Omega
To be converted into a voltmeter of range 0240 - 24 V there needs to be a resistance of 5988Ω5988\Omega

Note: While doing the calculations care must be taken to convert the unit of current from the given milliAmperes to Amperes. This conversion can be done by multiplying the milli Ampere value by 103{10^{ - 3}}, this will give the final value in Amperes. If the values are not converted to the SI units or if all the values are not in the same unit then our calculations can go wrong. The advantage of a galvanometer is that they have high sensitivity, high accuracy and it is not affected by stray magnetic fields. However, it can be used only for direct currents and result in errors due to damage of spring from mechanical stress.