CircuitNote

Quick electronics calculators for lab work and circuit design. Free, no sign-up, works in your browser.

Why your ammeter reads less current than you calculated

Splitting up the causes with real measurements from a series circuit.

Put a 1 kΩ and a 3 kΩ resistor in series, work out the current with Ohm's law, and the ammeter almost always reads a little less. The numbers below come from a real lab session. The measured current was 2.3 to 2.7% below the calculated value, and the measurements themselves show where the difference came from.

A R1 = 1 kΩ R23 kΩ +Vdc ammeter (internal RA) VR1 VR2 I
The test circuit. Dashed lines show where the voltages were measured.
In this guide 7

1. Measurements

Supply settingVdcVR1VR2Ammeter I
Supply setting10 VVdc10.023 VVR12.480 VVR27.503 VAmmeter I2.4381 mA
Supply setting16 VVdc16.015 VVR13.963 VVR211.856 VAmmeter I3.9119 mA

The resistors are marked 1 kΩ and 3 kΩ (±5%). Voltages and current were read on the same multimeter.

2. Compared with the calculation

SupplyVdc ÷ 4 kΩVR1 ÷ 1 kΩVR2 ÷ 3 kΩAmmeterShortfall
Supply10 VVdc ÷ 4 kΩ2.5057 mAVR1 ÷ 1 kΩ2.480 mAVR2 ÷ 3 kΩ2.501 mAAmmeter2.4381 mAShortfall−2.7%
Supply16 VVdc ÷ 4 kΩ4.0038 mAVR1 ÷ 1 kΩ3.963 mAVR2 ÷ 3 kΩ3.952 mAAmmeter3.9119 mAShortfall−2.3%

Against the supply voltage divided by the marked 4 kΩ, the ammeter reads −2.7% at 10 V and −2.3% at 16 V. Using the voltage across each resistor instead, it is still 1.0 to 2.5% low. You can check these differences with the percent error calculator.

3. Splitting the causes: Vdc ÷ I

Dividing the supply voltage by the ammeter reading gives the resistance the whole loop actually had. The same current flows through R1 and R2, so each resistor's voltage divided by I gives its real value. Whatever is left over sits outside the resistors, in the ammeter, the leads and the contacts.

SupplyVdc ÷ I (whole loop)R1 actualR2 actualOutside the resistors (meter, leads)
Supply10 VVdc ÷ I (whole loop)4,111 ΩR1 actual1,017 ΩR2 actual3,077 ΩOutside the resistors (meter, leads)16 Ω
Supply16 VVdc ÷ I (whole loop)4,094 ΩR1 actual1,013 ΩR2 actual3,031 ΩOutside the resistors (meter, leads)50 Ω

The loop behaved like about 4,094 to 4,111 Ω rather than 4 kΩ. The extra resistance splits into two parts.

  • The real resistor values: R1 + R2 was +1.1% to +2.4% above the marked value. That accounts for 85% of the extra resistance at 10 V and 47% at 16 V.
  • The ammeter and wiring: 0.040 V and 0.196 V were lost outside the resistors, which is 16 Ω and 50 Ω (0.4% and 1.3% of the total). That is the rest.

4. Resistor tolerance: ±5% is wider than it sounds

Measuring a few resistors from the same lab kit directly:

MarkedPart 1Part 2Part 3
Marked100 ΩPart 1101.35 Ω (+1.3%)Part 298.86 Ω (−1.1%)Part 398.12 Ω (−1.9%)
Marked1 kΩPart 11.0435 kΩ (+4.4%)Part 21.0212 kΩ (+2.1%)Part 3981.4 Ω (−1.9%)
Marked10 kΩPart 110.1 kΩ (+1.0%)Part 210.086 kΩ (+0.9%)Part 310.041 kΩ (+0.4%)
Marked100 kΩPart 198.98 kΩ (−1.0%)Part 2100.52 kΩ (+0.5%)Part 398.55 kΩ (−1.4%)

All are within ±5%, but the furthest is 4.35% off. The three 1 kΩ parts alone spread from −1.9% to +4.4%, and R1 in this circuit (about 1,017 Ω) falls in the same range. The resistor color code calculator shows the tolerance range for any marking.

5. Ammeter internal resistance (burden voltage)

An ammeter goes in series with the circuit, and inside it is a small shunt resistor it uses to sense the current. Connecting it adds RA to the loop, so the current drops.

I = (Vdc)/(R1 + R2 + RA)

RA ranges from a few ohms to a few hundred ohms depending on the meter and range. Here, together with the leads and contacts, it was in the tens of ohms. The smaller the circuit resistance, the bigger the effect: in a 100 Ω circuit the same RA could cut the current by more than 10%. Datasheets usually list it as the burden voltage.

Try it: split the missing current

It starts from the 10 V run: the actual R1 and R2 and the resistance outside them (meter and leads). Put each back to nominal and see where the gap comes from.

Current against supply voltage (dots: measured)

Calculated (4 kΩ nominal)
2.506 mA
Model current
2.438 mA
Against the calculation
−2.70%
Share from the resistors
85%
Share from the meter and leads
15%
Drop across meter and leads
40 mV

Things to try: drop the meter and lead resistance to 0 and whatever gap is left belongs to the resistor tolerance. Put both resistors back to 1 kΩ and 3 kΩ and only the meter's share remains.

6. The meter's own accuracy

R2 works out to 3,077 Ω at one supply setting and 3,031 Ω at the other, 1.5% apart. The resistor itself does not change that much (R2 dissipates only about 46 mW, so heating is negligible). A spread like this is within the multimeter's own accuracy. A typical handheld meter is specified around ±(0.5% + a few digits) for DC volts and around ±(1% + a few digits) for DC current; check the manual for your model. The multimeter accuracy calculator turns a spec into an uncertainty.

7. In your lab report

  • Measure each resistor with the multimeter before building, and use the measured values instead of the marked ones in your theory calculation. That removes the resistor share of the gap.
  • Measure the voltages with the ammeter still in the circuit. Then Vdc − VR1 − VR2 is the voltage across the ammeter.
  • Example discussion: "The measured current was 2.7% below the theoretical value. The loop resistance, Vdc divided by I, was 4,111 Ω, which we attribute to the real resistor values (±5% tolerance) plus the ammeter's internal resistance."

Check the Ohm's law step with the Ohm's law calculator and the series total with the series and parallel calculator. The function generator and oscilloscope half of the same lab is in the oscilloscope guide.