Lesson 3: Analog Measuring Instruments - Galvanometer, Ammeter, Voltmeter and Ohmmeter

• Detects very weak currents, measures their intensity and determines their direction (polarity).
• It works on the torque on a current-carrying coil in a magnetic field.

• A rectangular coil of many turns of thin insulated copper wire, wound on a light aluminium frame.
• The frame is mounted on a soft-iron cylinder (core) between the poles of a strong permanent horseshoe magnet; the pole pieces are concave (curved) so that the field in the gap is radial.
• The coil is pivoted on jewel (agate) bearings and carries a pointer moving over a scale (zero at the centre in a centre-zero instrument).
• Two spiral (hair) springs, above and below, provide the restoring torque and also act as current leads to the coil.

1. The current in the coil gives a couple τ = BIAN. The coil turns until this torque is balanced by the opposite torque of the twisted springs, which is proportional to the angle of deflection (τ_spring = c θ, where c is the torsion constant of the springs).
2. The radial field and soft-iron core keep the plane of the coil always parallel to the field lines, so the sides always cut the field at 90° and B is the same magnitude at every angle. Then the deflecting torque BIAN is independent of the angle, and the balanced angle θ is proportional to the current → the scale is uniform.
3. Reversing the current reverses the deflection.

At balance: B I A N = c θ, so θ = (B A N / c) I. The factor BAN/c is the sensitivity.

• Soft-iron core: concentrates the flux, strengthens the field in the gap and, with the concave poles, makes the field radial.
• Spiral springs: give a restoring torque proportional to the angle (so equilibrium is reached), and carry the current to the coil without hindering its rotation. They also bring the pointer back to zero when the current stops.
• Light aluminium frame: low inertia and damps the oscillations by eddy currents so the pointer settles quickly.
• Concave pole pieces: make the field radial so the scale is uniform.
• Many turns of thin wire and a strong magnet: large torque for a small current (high sensitivity).
• Permanent magnet (not an electromagnet): the field must be constant and independent of any current.

• The sensitivity of a galvanometer is the deflection (in degrees or scale divisions) produced per unit current through the coil, e.g.
• degree/μA.
• It increases with a large number of turns, a larger coil area and a stronger field, and with a weaker control spring (smaller c).

Example 7a. A galvanometer deflects by 40° when the current is 20 μA. Sensitivity = 40/20 = 2 degree/μA. Because the deflection is proportional to the current, a current of 35 μA gives 70°, and a deflection of 25° indicates a current of 12.5 μA.

• An ammeter measures current directly and is connected in series in the circuit.
• A galvanometer coil can carry only a very small current I_g (full-scale deflection, FSD).
• To measure a larger current I the instrument is converted: a very small resistance, the shunt R_s, is connected in parallel with the galvanometer coil (resistance R_g).

Reasoning: the main current I divides into I_g through the coil and I_s through the shunt, so I_s = I − I_g. The shunt and the coil have the same potential difference (they are in parallel):

I_g R_g = (I − I_g) R_s

R_s = I_g R_g / (I − I_g)

Notes:

• The shunt is small, so the combined resistance of the ammeter R_g R_s/(R_g + R_s) is very small, which does not change the current in the circuit when connected in series. In fact this resistance equals I_g R_g / I.
• Most of the current passes through the shunt; only a fraction I_g/I passes through the coil.
• A larger measuring range needs a smaller shunt. To extend the range n times (I = n I_g): R_s = R_g / (n − 1).
• The scale is recalibrated so that full scale reads I, not I_g.
• The pointer in a DC ammeter deflects in one direction, so the positive terminal of the ammeter must be joined to the side nearer the positive pole of the source.

Example 7. A galvanometer of resistance 4 Ω gives FSD at 10 mA. To read up to 2 A: R_s = 0.010 × 4 / (2 − 0.010) = 0.04/1.99 = 0.0201 Ω.

Example 7c. A galvanometer has R_g = 60 Ω and FSD current 1.5 mA. To convert it into an ammeter of range 0.3 A: R_s = 0.0015 × 60 / (0.3 − 0.0015) = 0.09 / 0.2985 = 0.3015 Ω. Total resistance of the ammeter = 60 × 0.3015 / 60.3015 = 0.300 Ω (= I_g R_g / I = 0.09/0.3 ✔).

Example 7d — range ×10. A galvanometer of 36 Ω is to read ten times its FSD current: R_s = 36 / (10 − 1) = 4 Ω in parallel. Check: when the main current is 10 I_g, 9 I_g goes through 4 Ω giving 36 I_g, and I_g goes through 36 Ω giving 36 I_g ✔.

A voltmeter measures potential difference and is connected in parallel across the part of the circuit concerned. The voltage across a galvanometer coil at FSD is tiny (V_g = I_g R_g), so the bare galvanometer could only measure very small voltages. To measure larger voltages a large resistance, the multiplier R_m, is connected in series with the coil.

At FSD the instrument must carry I_g when the voltage across the whole instrument is V:

V = I_g R_g + I_g R_m → R_m = V / I_g − R_g (equivalently R_m = (V − V_g)/I_g)

Notes:

• The total resistance of the voltmeter (R_g + R_m) is very large, so it draws a negligible current and does not change the voltage being measured.
• The positive terminal of the voltmeter is connected to the point of higher potential; if the voltage is reversed, the connections must be reversed.
• The multiplier makes the resistance large, so most of the measured voltage is across the multiplier and only a tiny voltage (I_g R_g) is across the coil.

Example 8. Galvanometer: R_g = 20 Ω, I_g = 5 mA. To read up to 30 V: R_m = 30/0.005 − 20 = 6000 − 20 = 5980 Ω in series.

Example 8a. R_g = 40 Ω, I_g = 1 mA, desired range 12 V: R_m = 12/0.001 − 40 = 11 960 Ω. Total resistance of the voltmeter = 12 000 Ω. For a 3 V range the multiplier would be 3000 − 40 = 2960 Ω.

Example 8b. What is the effect of the voltmeter resistance? A voltmeter of 12 000 Ω is connected across a 4000 Ω resistor in a circuit. The combination 4000 × 12 000 / 16 000 = 3000 Ω, so the voltmeter lowers the resistance between the two points by 25%, which is why a voltmeter's resistance must be as large as possible compared with the resistance across which it is placed.

An ohmmeter measures an unknown resistance directly. Measuring a resistance by Ohm's law needs both an ammeter and a voltmeter (R = V / I). If the voltage is fixed and known, we can drop the voltmeter and mark the galvanometer scale in ohms instead of amperes.

It is a micro-ammeter (galvanometer) in series with a fixed resistance, a variable resistance (adjusted for zero) and a cell of constant emf (about 1.5 V), with two terminals between which the unknown resistance R_x is connected.

• Zero adjustment: with the terminals touching (R_x = 0, short circuit), the variable resistor is adjusted so that the pointer gives full-scale deflection: R_total = V_B / I_g.
• With R_x inserted the current is I = V_B/(R_total + R_x), which is smaller, so the deflection is smaller and the scale is calibrated directly in ohms.
• Features of the scale: zero ohms is at the full-scale end (the right) and infinite resistance at the left (open circuit) — opposite to the current scale; it is non-linear (the readings are spread out at the zero end and crowded at the high-resistance end).
• If R_x = R_total the current is half of FSD; if R_x = 2 R_total it is one-third FSD, etc. In general, I / I_g = R_total / (R_total + R_x).
• Reasons: the cell must have a constant emf (otherwise the calibration is wrong); the fixed resistor protects the meter and sets the calibration; the variable resistor is needed to bring the pointer exactly to the zero mark when the cell ages.

Example 9. An ohmmeter has a 1.5 V cell and a total internal resistance of 3000 Ω at FSD (500 μA). With an unknown resistor of 3000 Ω: I = 1.5/6000 = 250 μA = half of FSD. With 9000 Ω: I = 1.5/12000 = 125 μA = quarter of FSD.

Example 9a — a calibration table. An ohmmeter has a 1.5 V cell and FSD current 1 mA, so its total internal resistance is 1.5/0.001 = 1500 Ω. For different R_x:

Example 9b — design. A milliammeter of resistance 20 Ω has FSD 10 mA. It is used as an ohmmeter with a 1.5 V cell of internal resistance 1 Ω. For FSD the total resistance must be 1.5/0.010 = 150 Ω, so the standard resistor needed (including the variable one) is 150 − 20 − 1 = 129 Ω. A reading of 6 mA means the total resistance is 1.5/0.006 = 250 Ω, so R_x = 250 − 150 = 100 Ω. With R_x = 600 Ω the current is 1.5/750 = 2 mA.

• Current: convert it into an ammeter by a shunt and connect it in series.
• emf (or potential difference): convert it into a voltmeter by a multiplier and connect it in parallel across the source. For the emf the source must be on open circuit (no current flowing), with a voltmeter of very high resistance.
• Resistance: convert it into an ohmmeter, or use an ammeter and a voltmeter together to find R = V/I.

• Analogue (pointer) meters use a pointer on a scale; a multimeter can be switched to measure current, voltage and resistance.
• Digital multimeters show the value as numerals on an LCD (liquid-crystal display) without a pointer.

• A soft-iron cylinder is fixed inside the coil of the galvanometer: to strengthen the field and make it radial.
• The coil is attached to a pair of spiral springs: to supply a restoring torque proportional to the angle and to lead the current to the coil.
• A high resistance is connected in series with the galvanometer when it is used as a voltmeter: to limit the current to I_g at FSD and to make the voltmeter draw negligible current.
• An ammeter is connected in series and a voltmeter in parallel: so that the ammeter carries the whole current of the branch and the voltmeter sees the same voltage as the element.
• A constant resistor is placed inside the ohmmeter: to protect the meter and fix the calibration.
• The cell of an ohmmeter must have a constant emf: a changing emf would make the calibration wrong.
• The scale of the ohmmeter is not linear and runs in the opposite sense to that of the ammeter: because I = V_B/(R_total + R_x) is not proportional to R_x, and larger resistance gives smaller current.