Converting One Galvanometer into an Ammeter and a Voltmeter
A galvanometer has a full-scale deflection current of 1.00 mA and a coil resistance of 100 Ω. A student wants to make two meters from it: a 0–1.00 A ammeter and a 0–5.00 V voltmeter. The available resistors are 0.10 Ω at 5 W, 4.9 kΩ at 0.25 W, and 100 Ω at 0.25 W. The student first plans to put the 0.10 Ω resistor in series with the galvanometer for the voltmeter and the 4.9 kΩ resistor in parallel for the ammeter, saying the small resistor will protect the meter and the large one will shunt most of the current. The teacher says the connections are reversed. The ammeter will be used in a circuit with a 6.0 V battery and a 12 Ω resistor. The voltmeter will be used to measure the voltage across one of two 4.7 kΩ resistors connected in series across an 8.0 V supply. The student must choose the correct connections and decide whether each measurement is trustworthy.
1. Calculate the correct shunt resistance for the 1.00 A ammeter and the correct series resistance for the 5.00 V voltmeter. Compare with the available resistors and state which resistor goes where.
Read the worked answer
For an ammeter, the shunt S is in parallel with the galvanometer and carries most of the current:
I_gR_g = (I − I_g)S S = I_gR_g/(I − I_g) = (1.00×10⁻³)(100)/(1.00 − 1.00×10⁻³) = 0.1001 Ω.
The available 0.10 Ω resistor is therefore the correct shunt. With it, full-scale current is I = I_g(1 + R_g/S) = 0.001(1 + 100/0.10) = 1.001 A, close enough to 1.00 A.
For a voltmeter, the series resistor R_s must drop the excess voltage at full-scale current:
V = I_g(R_g + R_s) R_s = V/I_g − R_g = 5.00/0.001 − 100 = 4900 Ω.
The 4.9 kΩ resistor is exactly the required series resistor. The 100 Ω resistor is not needed in either meter. The student’s planned connections are reversed: the small resistor must be in parallel for the ammeter, and the large resistor in series for the voltmeter.
2. Check power dissipation in the chosen resistors at full-scale deflection and comment on whether the 5 W and 0.25 W ratings are adequate.
Read the worked answer
In the ammeter shunt, the current at full scale is about 1.00 A, so P ≈ I²S = (1.00)²(0.10) = 0.10 W. This is far below the 5 W rating.
In the voltmeter series resistor, the current is only the galvanometer full-scale current, 1.00 mA. Thus P = I_g²R_s = (1.00×10⁻³)²(4900) = 4.9×10⁻³ W, far below 0.25 W. The galvanometer coil itself dissipates only I_g²R_g = 1.0×10⁻⁵ W.
Both chosen resistors are adequately rated. The 5 W rating of the shunt is generous; a smaller rating would still work, but the high rating gives a safety margin if the meter is briefly overloaded.
3. Predict the ammeter reading in the 6.0 V, 12 Ω circuit and the voltmeter reading across one 4.7 kΩ resistor in the 8.0 V series circuit. Compare each with the value that would be obtained with ideal meters, and decide which measurement is acceptable.
Read the worked answer
Ammeter: its resistance is about 0.10 Ω, so the total circuit resistance is 12.10 Ω. The current is I = 6.0/12.10 = 0.496 A. An ideal ammeter would read 6.0/12.0 = 0.500 A. The error is about 0.8%, so this measurement is trustworthy for most school purposes.
Voltmeter: its resistance is R_V = R_g + R_s = 100 + 4900 = 5000 Ω. When connected across one 4.7 kΩ resistor, the meter and that resistor are in parallel:
R_p = (4700)(5000)/(4700 + 5000) = 2423 Ω.
The total series resistance becomes 2423 + 4700 = 7123 Ω. The voltage across the measured resistor is V = 8.0(2423/7123) = 2.72 V. The ideal value would be 4.00 V, so the loading error is about 32% - not acceptable.
The ammeter measurement is acceptable; the voltmeter measurement is not, because a 5 kΩ voltmeter is comparable to the 4.7 kΩ circuit resistance. A higher-resistance meter, such as a digital voltmeter with 10 MΩ input resistance, would be needed.