Lab guide

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.

1. Measurements

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

2. Compared with the calculation

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: V_dc ÷ 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.

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:

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 R_A to the loop, so the current drops.

I = V_dc / (R1 + R2 + R_A)

R_A 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 R_A could cut the current by more than 10%. Datasheets usually list it as the burden voltage.

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 V_dc − V_R1 − V_R2 is the voltage across the ammeter.
• Example discussion: "The measured current was 2.7% below the theoretical value. The loop resistance from V_dc/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.