Converting One Galvanometer into an Ammeter and a Voltmeter
A laboratory galvanometer has a resistance of 50 Ω and gives full-scale deflection at 1.0 mA. A student is asked to convert it into (i) a 0–1 A ammeter and (ii) a 0–10 V voltmeter, using the same movement. She calculates a shunt of about 0.05 Ω and a series resistance of about 9950 Ω and mounts both in one box with a selector switch. On testing, the ammeter reads correctly at 0.5 A. When the box is switched to the voltmeter range and connected across one of two equal 10 kΩ resistors in series across a 20 V supply, the reading is 6.7 V instead of the expected 10 V. In a second test, the ammeter is inserted in series with a 0.10 Ω strip carrying about 1 A, and its reading is noticeably lower than that of a clamp meter. The teacher asks the group to explain both observations in terms of instrument resistance and loading, and to state whether the meters are faulty.
1. Verify the shunt and series resistance values and state the resistance of each finished instrument.
Read the worked answer
Given G = 50 Ω and I_g = 1.0 mA = 1.0 × 10⁻³ A.
Ammeter (0–1 A) The shunt S is in parallel with the galvanometer and must carry the excess current: I − I_g = 1.0 − 0.001 = 0.999 A. The p.d. across the parallel combination at full scale is V_g = I_gG = (1.0 × 10⁻³)(50) = 0.050 V. So S = V_g/(I − I_g) = 0.050/0.999 = 0.05005 Ω ≈ 0.050 Ω. The ammeter resistance is R_A = GS/(G + S) = (50)(0.050)/(50.050) ≈ 0.050 Ω. Most of the current bypasses the coil; only 1 mA passes through it at full scale.
Voltmeter (0–10 V) The series resistance R_s must drop the excess voltage when 1 mA flows: Total resistance needed = V/I_g = 10/(1.0 × 10⁻³) = 10,000 Ω. R_s = 10,000 − 50 = 9,950 Ω. The finished voltmeter has a resistance of 10 kΩ, and it draws 1 mA at full scale.
Both calculations are correct; the problem is not the arithmetic but the loading these resistances produce in real circuits.
2. Explain why the voltmeter reads 6.7 V instead of 10 V across the 10 kΩ resistor. Is the meter faulty?
Read the worked answer
The circuit is two 10 kΩ resistors in series across a 20 V supply. Without the meter, each takes 10 V.
When the 10 kΩ voltmeter is connected across one resistor, it is in parallel with that resistor: R_parallel = (10 kΩ × 10 kΩ)/(10 kΩ + 10 kΩ) = 5 kΩ. The other 10 kΩ resistor remains in series, so the total resistance becomes R_total = 5 kΩ + 10 kΩ = 15 kΩ. The current is I = 20/15,000 = 1.33 × 10⁻³ A, and the p.d. across the parallel section is V = IR_parallel = (1.33 × 10⁻³)(5,000) = 6.67 V.
This is the loading effect. A voltmeter is not ideal: its 10 kΩ resistance is comparable with the 10 kΩ resistor, so it draws a significant current and changes the very voltage it is measuring. The meter is not faulty; it is behaving as a 10 kΩ resistor. An ideal voltmeter would have infinite resistance. To reduce the error, use a voltmeter of much higher resistance (e.g. a digital meter with 10 MΩ input), or calculate and correct for the loading. The same movement would also read low in any high-resistance circuit; the higher the circuit resistance compared with the meter resistance, the worse the error.
3. Explain the low ammeter reading in the 0.10 Ω strip test. Is the ammeter faulty, and how should the measurement be improved?
Read the worked answer
The ammeter has R_A ≈ 0.050 Ω and must be placed in series with the strip. The total resistance of the current path therefore changes from about 0.10 Ω to R_total = 0.10 + 0.050 = 0.15 Ω.
If the rest of the circuit resistance is negligible, the current for the same driving voltage falls by the factor 0.10/0.15 = 0.667, i.e. about one-third lower - for example, from roughly 1.0 A to about 0.67 A. The clamp meter reads the true current because it does not insert resistance into the circuit.
The ammeter is not faulty; it is a low-resistance instrument, but 'low' is not 'zero'. This is the ammeter loading effect, and it is most serious when the load resistance is comparable with or smaller than the meter resistance.
Improvements: use an ammeter with a smaller shunt (lower R_A), use a clamp meter or Hall-effect probe that does not break the circuit, or measure the voltage across a known series resistor and calculate I = V/R. The general rule is that an ammeter must have resistance much smaller than the rest of the circuit, just as a voltmeter must have resistance much larger. Note also that an ammeter must never be connected in parallel with a load: it would short the load and draw a damaging current.