From galvanometer to ammeter, voltmeter and ohmmeter
Answers are for parents and teachers.
1
A galvanometer has Rg = 30 Ω and Ig = 5 mA. Find the shunt for a range of 3 A and the resistance of the ammeter.
Rs = 0.005 × 30 ÷ (3 − 0.005) = 0.15 ÷ 2.995 = 0.0501 Ω. Ammeter resistance = IgRg ÷ I = 0.15 ÷ 3 = 0.05 Ω.
2
The same galvanometer (Rg = 30 Ω, Ig = 5 mA) is made into a voltmeter. Find the multiplier for a 15 V range and for a 150 V range.
15 V: Rm = 15 ÷ 0.005 − 30 = 3000 − 30 = 2970 Ω. 150 V: Rm = 30 000 − 30 = 29970 Ω.
3
An 8 kΩ resistor and a 12 kΩ resistor are in series with a 20 V cell of negligible internal resistance. A voltmeter of resistance 12 kΩ is connected across the 12 kΩ resistor. Find the voltage across it before the voltmeter is connected and what the voltmeter reads.
Before: I = 20 ÷ 20 000 = 1 mA, V = 1 × 10⁻³ × 12 000 = 12 V. With the voltmeter: 12 kΩ in parallel with 12 kΩ is 6 kΩ, the total is 14 kΩ, I = 20 ÷ 14 000 = 1.43 mA, and the reading is 1.43 × 10⁻³ × 6000 = 8.57 V.
4
An ohmmeter has E = 1.5 V and a full-scale current of 1.0 mA. Find Ri, the current for Rx = 1500 Ω and for Rx = 4500 Ω, and Rx when the pointer rests at 0.2 mA.
Ri = 1.5 ÷ 0.001 = 1500 Ω. Rx = 1500 Ω: I = 1.5 ÷ 3000 = 0.5 mA. Rx = 4500 Ω: I = 1.5 ÷ 6000 = 0.25 mA. At 0.2 mA: Rx = 1.5 ÷ 0.0002 − 1500 = 7500 − 1500 = 6000 Ω.
5
Error hunt. A student wants an ammeter of range 0.25 A from a milliammeter with Rg = 5 Ω and Ig = 50 mA, and calculates the shunt as IgRg ÷ I = 1.0 Ω. Find the correct shunt and the range that the student’s shunt would really give.
The shunt carries only I − Ig, not I. Correct: Rs = 0.05 × 5 ÷ (0.25 − 0.05) = 0.25 ÷ 0.2 = 1.25 Ω. With Rs = 1.0 Ω the shunt current at full scale is 0.05 × 5 ÷ 1.0 = 0.25 A, so the range is 0.25 + 0.05 = 0.3 A, not 0.25 A.