Converting a Galvanometer into an Ammeter and a Voltmeter
In a school laboratory, a moving-coil galvanometer has a coil resistance of 40 Ω and gives full-scale deflection at 2.0 mA. A student is asked to convert this galvanometer into (i) a 0–5 A ammeter and (ii) a 0–10 V voltmeter, using only suitable resistors and the same galvanometer. The student first tries a shunt of 0.10 Ω for the ammeter and finds that the pointer goes off scale before the circuit current reaches 5 A. For the voltmeter, the student proposes a 5.0 kΩ series resistor and checks it by connecting the instrument across a 10 V supply. Later, while testing the ammeter, the student accidentally connects it in parallel with a 10 Ω resistor across a 2.0 V laboratory supply of negligible internal resistance. The teacher stops the test immediately and asks for a quantitative explanation of what would happen.
1. Calculate the correct shunt resistance for the 0–5 A ammeter and the correct series resistance for the 0–10 V voltmeter. Explain why the student's 0.10 Ω shunt made the ammeter go off scale too early.
Read the worked answer
For full-scale current I = 5 A, the shunt S carries (I − I_g) while the galvanometer carries I_g = 2.0×10⁻³ A. Since they are in parallel: I_g G = (I − I_g)S.
S = I_g G/(I − I_g) = (2.0×10⁻³)(40)/(5 − 2.0×10⁻³) = 0.080/4.998 = 1.60×10⁻² Ω ≈ 0.016 Ω.
With S = 0.10 Ω, full-scale current would be I = I_g(1 + G/S) = 2.0×10⁻³(1 + 40/0.10) = 0.80 A. So the pointer reaches full scale at only 0.80 A and goes off scale long before 5 A.
Voltmeter: R_s = V/I_g − G = 10/(2.0×10⁻³) − 40 = 5000 − 40 = 4960 Ω.
2. Find the resistance of the converted ammeter and voltmeter. Why must an ammeter have very low resistance and a voltmeter very high resistance?
Read the worked answer
Converted ammeter resistance: R_A = GS/(G + S) = (40)(0.016)/(40.016) ≈ 0.016 Ω, essentially the shunt. Converted voltmeter resistance: R_V = G + R_s = 40 + 4960 = 5000 Ω.
An ammeter is inserted in series, so it must not appreciably reduce the current; hence R_A must be very small. A voltmeter is connected in parallel, so it must draw negligible current from the component; hence R_V must be very large compared with the circuit resistance.
3. Predict what happens when the 0–5 A ammeter is connected in parallel with the 10 Ω resistor across the 2.0 V supply. Estimate the current through the ammeter and through the 10 Ω resistor and state why this is unsafe.
Read the worked answer
The ammeter's 0.016 Ω is almost a short circuit. In parallel with 10 Ω, R_eq = (0.016×10)/(10.016) ≈ 0.016 Ω. The supply current would be I = 2.0/0.016 ≈ 125 A, and almost all of it passes through the ammeter: I_A ≈ 2.0/0.016 ≈ 125 A; the 10 Ω resistor carries only 2.0/10 = 0.20 A.
The ammeter is being asked to read about 25 times its 5 A full-scale value. It may burn out, and the supply is effectively short-circuited. This is why an ammeter must never be connected in parallel with a low-resistance supply.