4.11 THE MOVING COIL GALVANOMETER

Currents and voltages in circuits have been discussed extensively in Chapters 3. But how do we measure them? How do we claim that current in a circuit is $1.5 A$ or the voltage drop across a resistor is $1.2 V$ ? Figure 4.24 exhibits a very useful instrument for this purpose: the moving coil galvanometer (MCG). It is a device whose principle can be understood on the basis of our discussion in Section 4.10.

The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis (Fig. 4.24), in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it. This torque is given by Eq. (4.26) to be $τ = N I A B$

where the symbols have their usual meaning. Since the field is radial by design, we have taken $s i n θ = 1$ in the above expression for the torque. The magnetic torque NIAB tends to rotate the coil. A spring $S_{p}$ provides a counter torque $k ϕ$ that balances the magnetic torque NIAB; resulting in a steady angular deflection $ϕ$ . In equilibrium $k ϕ = N I A B$

where $k$ is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection $ϕ$ is indicated on the scale by a pointer attached to the spring. We have

$ϕ = (\frac{N A B}{k}) I$ (4.38)

The quantity in brackets is a constant for a given galvanometer.

4.24

The moving coil galvanometer. Its elements are described in the text. Depending on the requirement, this device can be used as a current detector or for measuring the value of the current (ammeter) or voltage (voltmeter).

The galvanometer can be used in a number of ways. It can be used as a detector to check if a current is flowing in the circuit. We have come across this usage in the Wheatstone’s bridge arrangement. In this usage the neutral position of the pointer (when no current is flowing through the galvanometer) is in the middle of the scale and not at the left end as shown in Fig.4.24. Depending on the direction of the current, the pointer’s deflection is either to the right or the left.

The galvanometer cannot as such be used as an ammeter to measure the value of the current in a given circuit. This is for two reasons: (i) Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of $μ A$ . (ii) For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit. To overcome these difficulties, one attaches a small resistance $r_{s}$ , called shunt resistance, in parallel with the galvanometer coil; so that most of the current passes through the shunt. The resistance of this arrangement is, $R_{G} r_{s} / (R_{G} + r_{s}) ≃ r_{s} if R_{G} >> r_{s}$

If $r_{s}$ has small value, in relation to the resistance of the rest of the circuit $R_{c}$ , the effect of introducing the measuring instrument is also small and negligible. This arrangement is schematically shown in Fig. 4.25. The scale of this ammeter is calibrated and then graduated to read off the current value with ease. We define the current sensitivity of the galvanometer as the deflection per unit current. From Eq. (4.38) this current sensitivity is,

$\frac{ϕ}{I} = \frac{N A B}{k}$ (4.39)

4.25

Conversion of a galvanometer (G) to an ammeter by the introduction of a shunt resistance

r_{s}

of very small value in parallel.

A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns $N$ . We choose galvanometers having sensitivities of value, required by our experiment.

The galvanometer can also be used as a voltmeter to measure the voltage across a given section of the circuit. For this it must be connected in parallel with that section of the circuit. Further, it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large. Usually we like to keep the disturbance due to the measuring device below one per cent. To ensure this, a large resistance $R$ is connected in series with the galvanometer. This arrangement is schematically depicted in Fig.4.26. Note that the resistance of the voltmeter is now, $R_{G} + R ≃ R : large$

4.26

Conversion of a galvanometer (G) to a voltmeter by the introduction of a resistance

R

of large value in series.

The scale of the voltmeter is calibrated to read off the voltage value with ease. We define the voltage sensitivity as the deflection per unit voltage. From Eq. (4.38),

$\frac{ϕ}{V} = (\frac{N A B}{k}) \frac{I}{V} = (\frac{N A B}{k}) \frac{1}{R}$ (4.40)

An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Eq. (4.39) which provides a measure of current sensitivity. If $N \to 2 N$ , i.e., we double the number of turns, then $\frac{ϕ}{I} \to 2 \frac{ϕ}{I}$

Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.40), $N \to 2 N$ , and $R \to 2 R$ , thus the voltage sensitivity, $\frac{ϕ}{V} \to \frac{ϕ}{V}$ remains unchanged. So in general, the modification needed for conversion of a galvanometer to an ammeter will be different from what is needed for converting it into a voltmeter.

Example 4.13

In the circuit (Fig. 4.27) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance $R_{G} = 60.00 Ω$ ;
(b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance $r_{s} = 0.02 Ω$ ;
(c) is an ideal ammeter with zero resistance?

4.27

VIEW SOLUTION

Total resistance in the circuit is, $R_{G} + 3 = 63 Ω$ Hence, $I = 3 / 63 = 0.048 A$
Resistance of the galvanometer converted to an ammeter is, $\frac{R_{G} r_{s}}{R_{G} + r_{s}} = \frac{60 Ω \times 0.02 Ω}{(60 + 0.02) Ω} ≃ 0.02 Ω$ Total resistance in the circuit is, $0.02 Ω + 3 Ω = 3.02 Ω$ Hence, $I = 3 / 3.02 = 0.99 A$
For the ideal ammeter with zero resistance, $I = 3 / 3 = 1.00 A$