Galvanometer – Types, Moving Coil, Transformation and Solved Examples

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What is Galvanometer and Voltmeter? What are the Disadvantages of Using a Galvanometer?

Galvanometer is an instrument used to detect and measure the current in a circuit.

A current measuring instrument can be constructed using the heating effect, the chemical effect or the magnetic effect of cur-rent. However to measure a direct current (dc), the magnetic effect is most advantageous for practical purposes.

The two types of galvanometers used widely in laboratory are—

i) Moving magnet galvanometer : The basic principle of this instrument is the action of electric current on a magnet.
Example: Tangent galvanometer, Since galvanometer, Helmholtz double coil galvanometer.
This type of galvanometers is rarely used due to its disad-vantageous setup and low sensitivity. The discussion is out of our syllabus.

ii) Moving coil galvanometer : The basic principle of this instrument is the action of a magnet on electric current.
Example: D’Arsonval galvanometer, table galvanometer.

Moving Coil Galvanometer

With a moving coil galvanometer, we can detect and measure even a very weak current (about 10^-9A ) in a circuit. This galva-nometer is shown in Fig.

Description of a Moving Coil:

ABCD : A coil of insulated copper wire wound over a rectangu-lar frame made of cane or aluminium [Fig.(a), (c)].

I: A cylinder of soft iron. The axis of the cylinder coincides with the axis of the coil and the cylinder is fitted inside the coil.

NS: Two magnetic poles made of soft iron attached with a strong permanent magnet. The gap between the two poles is cylindrical and in this gap the coil is placed in such a way that the axis of the gap coincides with the axis of the coil.

Due to this special shape of the magnetic poles [Fig.(b)], the magnetic field always remains parallel to the plane of the coil.

Now this coil is fitted with a mechanical system such that—

1. the circuit current to be measured can be sent directly through the coil and
2. very accurate measurement of deflection of the coil due to flow of current can be done.

For this purpose, two types of mechanical arrangements are widely used :

i) Suspended-coil galvanometer or D’Arsonval galvano-meter: This type of arrangement is shown in Fig.(a).

W: Phosphor-bronze wire. The coil ABCD is hung by this wire.

M: A small mirror attached with the phosphor-bronze thread W. If the coil rotates through an angle θ, the mirror also rotates through θ. So, if a ray of light falls on the mirror from a stationary source of light, the reflected ray rotates through an angle 2θ [Fig.(d)], If the displacement of the reflected light-spot on a scale, kept at a distance D from the mirror, be d then,
2θ = \(\frac{d}{D}\) (when, d \(\ll\)D)
or, θ = \(\frac{d}{2 D}\) ……. (1)
For this arrangement, it is also called a mirror galvanome-ter.

ii) Table galvanometer: In Fig.(c), this arrangement is shown. Two magnetic poles are not drawn here.

HH: Two hair springs. Two ends of the coil are connected with these two hair springs. The circuit current to be deter-mined enters the coil through the two bearings P and Q attached to HH.

R: A pointer. With the rotation of the coil, it also rotates. When no current flows through the galvanometer, the pointer remains vertical. It can move on either side of this

Voltmeter: The Instrument used for measuring the potential difference between two points of an electrical circuit is called a voltmeter. To equalise the potential difference between those two points with the potential difference between the two ends of the voltmeter, it should be connected in parallel curent with those two points in the circuit [Fig.]. Again, the resistance of the voltmeter should be very high so that no appreciable fraction of the main current passes through the voltmeter. So, a voltmeter (or millivoltmeter) is an instrument of high resistance connected in parallel with an electrical circuit. Resistance of an Ideal voltmeter should be infinite.

Disadvantages of using a galvanometer: The basic instrument used for measuring current is a moving magnet galvanometer or a moving coil galvanometer. In spite of this, there are some problems to use these instruments directly in laboratory experiments.

i) Some time is spent for levelling and setting of these galvanometers.

ii) The magnitude of current cannot be obtained directly from any scale. To calculate the magnitude of current we have to use the galvanometer formula after measuring the angle of deflection.

iii) Usually the resistance of a galvanometer is of intermediate order (in most of the cases, from 100Ω to 500Ω). When this galvanometer is used to measure current or potential difference, the results turn out to be erroneous; because it is not an ideal one to use as an ammeter as its resistance is not too low, and again ills not an ideal one as a voltmeter as its resistance is not too high.

Transformation of Galvanometers

To make a galvanometer fit for everyday use in the laboratory, it should be effectively converted into an ammeter or a voltmeter. Thus, a galvanometer is the primary instrument while an ammeter or a voltmeter are the secondary instruments.

To construct an ammeter or a voltmeter, a moving coil table galvanometer is generally used. For this

1. the resistance of the galvanometer, G; and
2. the current I_G, for the full-scale deflection of the pointer in the circular scale, should be known beforehand.

Galvanometer to ammeter: Let the resistance of the galvanometer = G and the current required for the full-scale deflection of the pointer = 1 [Fig.].

Now, a low resistance in parallel, i.e.. a shunt S is connected with the galvanometer. The whole system is kept in a box covered with glass in such a way that

1. the pointer and the circular scale remain visible from outside, and

2. the terminals A and B are outside the box. The points A and B are connected with the external circuit. If main circuit current I corresponds to a full-scale deflection of the galvanometer, the galvanometer is effectively converted into an ammeter suitable for measuring maximum current I. Note that, due to the parallel combination of G and S, the equivalent resistance of the ammeter thus formed becomes sufficiently small.

Calculation: In Fig.,
V_X – V_Y = I_G ᐧ G = I_S ᐧ S [here, I_S stands for shunt current]
So, S = \(\frac{I_G}{I_S} \cdot G\)
Again, I = I_G + I_S or, I_S = I – I_G
Hence, S = \(\frac{I_G}{I-I_G} \cdot G\) …….. (1)
So, to convert the galvanometer into an ammeter to measure a maximum current I, a shunt S as obtained from equation (1), should be connected in parallel with the galvanometer.

Range of the ammeter: In equation (1), I > I_G; hence when an ammeter is constructed from a galvanometer, its range always increases and never decreases. If this range becomes n times, \(\frac{I}{I_G}\) = n; so from equation (1) we get,
S = \frac{1}{I / I_G-1} \cdot G\(\) = \(\frac{G}{n-1}\) …… (2)

Resistance of the ammeter The equivalent resistance of G and S is the resistance (R) of the ammeter.
So, \(\frac{1}{R}\) = \(\frac{1}{G}+\frac{1}{S}\) = \(\frac{1}{G}+\frac{n-1}{G}\) = \(\frac{n}{G}\) i.e., R = \(\frac{G}{n}\) …. (3)

Galvanometer to voltmeter: In this case, a high resistance R is connected in series with the galvanometer [Fig.].

The two points A and B are connected with the external circuit. So, if the current in the external circuit be I_G, galvanometer current will also be I_G and as a result, the pointer gives full-scale deflection. Hence, in this condition, the potential difference (V_A – V_B) between the two points A and B will show full-scale deflection of the pointer. So, the galvanometer will then be effectively converted into a voltmeter. Due to the series combination of G and R, the equivalent resistance of the voltmeter becomes sufficiently large.

Calculation: in Fig.,
V_A – V_B = V
or, G + R = \(\frac{V}{I_G}\) ∴ R = \(\frac{V}{I_G}\) – G …. (4)
So, to convert the galvanometer into a voltmeter fit for measuring a maximum potential difference V, a high resistance of magnitude R, as obtained from equation (4), should be connected in series.

Range of voltmeter: In the equation (4), V > I_GG (=V_G) and hence if a galvanometer is converted into a voltmeter, the range of the voltage always increases and never decreases. If this range becomes n -times, then \(\frac{V}{V_G}\) = n; So from equation (4) we get,
R = \(G\left(\frac{V}{I_G \cdot G}-1\right)\) = \(G\left(\frac{V}{V_G}-1\right)\) = G(n – 1) …. (5)
Resistance of the voltmeter: Naturally, resistance of the newly formed voltmeter,
R_V = G + R = G + G(n – 1) = nG ….. (6)

Differences between an ammeter and a voltmeter:

Ammeter	Voltmeter
1. An ammeter is a current measuring instrument.	1. A voltmeter is a potential difference measuring instrument.
2. It is connected in series with a circuit.	2. It is connected in parallel with a circuit.
3. Internal Resistance is very low.	3. Internal Resistance is very high.
4. It is constructed by connecting a low-resistance shunt in parallel with a moving coil galvanometer.	4. It is constructed by connecting a high resistance in series with a moving coil galvanometer.

Theoretically moving magnet or suspended-coil galvanometers may be converted to ammeter or voltmeter in a similar way. But due to the practical difficulties associated with such arrangements only moving coil table galvanometers are converted to be used as ammeter or voltmeter.

Numerical Examples

Example 1.
A galvanometer of resistance 10Ω gives full scale deflection for a Current of 10 mA. How can this galvanometer be used
(i) as an ammeter to measure current of range 0-2 A and
(ii) as a voltmeter having voltage range 0-5 V? [HS’ 01]
Solution:
Resistance of the galvanometer, G = 10Ω; maximum current, I_G = 10 mA = 0.01 A

i) Connecting a shunt S in parallel to the galvanometer [Fig.], if the instrument is used between the points A and B, an ammeter of current-range 0-I will be obtained.
In the given question, I = 2 A.

ii) A voltmeter having voltage range of zero to (V_A – V_B) is obtained if a resistance R is connected in series with the galvanometer and the instrument is used between the points A and B [Fig.].

In the given problem, V_A – V_B = 5 V
Now, V_A – V_B = I_G(G + R)
or, R = \(\frac{V_A-V_B}{I_G}\) – G = \(\frac{5}{0.01}\) – 10 = 500 – 10 = 490Ω
This is the required value of the resistance R.

Example 2.
Full-scale deflection occurs in a moving coil galvanometer of resistance 36Ω when 100 mA current flows through it. What arrangement should be done to convert it into a voltmeter of 0 – 5 V range? Draw the necessary circuit diagram.
Solution:
Resistance of the galvanometer = 360, maximum current, I_G = 100 mA = 0.1 A .
Connecting a resistance R in series with the galvanometer and using the equipment in between the two points A and B, we shall get a voltmeter to measure voltage difference ranging from 0 to (V_A – V_B) [Fig.]. Here, V_A – V_B = 5 V
Now, V_A – V_B = I_G(G + R)
or, R = \(\frac{V_A-V_B}{I_G}\) – G = \(\frac{5}{0.1}\) – 36 = 50 – 36 = 14Ω
This is the required value of the resistance R.

Example 3.
A millivoltmeter of range ‘0-50 mV’ and resistance 500 is to be converted into an ammeter of range 0-1 A. How can it be done?
Solution:
Resistance of the millimeter, R = 50Ω ; maximum voltage, V = 50 mV = 0.05V
Connecting a shunt S in parallel with the millivoltmeter and using this combination in between the points A and B [Fig.], we shall get an ammeter to measure a maximum current I. Here, I = 1 A.

This is the required value of the shunt-resistance.

Example 4.
How would you convert a voltmeter that can measure up to 150 V to an ammeter which can measure current up to 8 A ? Resistance of the voltmeter is 300Ω.
Solution:
Resistance of the voltmeter, R = 300Ω; maximum voltage, V = 150V.
∴ Maximum current through the voltmeter,
I_V = \(\frac{V}{R}\) = \(\frac{150}{300}\) = 0.5 A
Connecting a shunt S in parallel with the voltmeter [Fig.] and using this combination in between the points A and B, we shall get an ammeter to measure a maximum current I.
Here, I = 8A
Now, I_S ᐧ S = I_V ᐧ R
or, S = \(\frac{I_V}{I_S} \cdot R\) = \(\frac{I_V}{I-I_V} \cdot R\) = \(\frac{0.5}{8-0.5} \times 300\) = \(\frac{0.5}{7.5} \times 300\)
= 20Ω
This is the required value of the shunt-resistance.

Example 5.
A galvanometer of resistance 100Ω gives full scale deflection for a current of 10 mA. What is the value of the shunt to be used to convert it into an ammeter which can measure current up to 10 A?
Solution:
Resistance of the galvanometer, G = 100Ω; maximum current, I_G = 10 mA = 0.01 A. Connecting a shunt S in parallel with the voltmeter [Fig.] and using this combination
in between the points A and B, we shall get an ammeter useful to measure a maximum current I.
Here, I = 10A and I_S ᐧ S = I_G ᐧ G
or S = \(\frac{I_G}{I_S} \cdot G\) = \(\frac{I_G}{I-I_G} \cdot G\) = \(\frac{0.01}{10-0.01} \times 100\) = \(\frac{1}{9.99}\)
= 0.1001Ω

Example 6.
A moving coil galvanometer of resistance 50Ω gives full scale deflection for a current of 50 mA. How can this galvanometer be used to convert it into a voltmeter which can measure voltage upto 200 V’
Solution:
Resistance of the galvanometer, G = 50Ω; maximum current, I_G = 50 mA = 0.05 A. Connecting a resistance R in series with (he galvanometer and using the equipment in between the two points A and B, we shall get a voltmeter to measure voltage difference ranging from 0 to (V_A – V_B) [Fig.]
Here, V_A – V_B = 200V
Now, V_A – V_B = I_G(G + R)
or R = \(\frac{V_A-V_B}{I_G}\) – G = \(\frac{5}{0.1}\) – 36 = 14Ω
This is required value of resistance R.