Two Resistors and a Non-Ideal Ammeter
A student builds a circuit with a 12 V battery of negligible internal resistance, a 6 Ω resistor and a 3 Ω resistor in series, and an ammeter inserted in the loop. The ammeter is not ideal: it has a resistance of 2 Ω. The student expects a current of 12/(6+3) = 1.33 A. A voltmeter (ideal) is then connected across the 3 Ω resistor. The student is surprised to read a current lower than expected and asks why the ammeter matters and whether the voltmeter reading equals 3 × I.
1. Calculate the actual current and explain the discrepancy with the student's expectation.
Read the worked answer
The ammeter is in series, so its resistance adds to the total: $R_{total} = 6 + 3 + 2 = 11\ \Omega$.
$I = \dfrac{12}{11} \approx 1.09$ A, not 1.33 A.
The student's error was treating the ammeter as a zero-resistance wire. A real ammeter has some resistance, and in series it reduces the current. An ideal ammeter would have zero resistance; the 2 Ω here is a 22 % increase over the 9 Ω the student assumed, so the current drops by roughly that fraction.
2. The ideal voltmeter is connected across the 3 Ω resistor. Predict its reading and justify why an ideal voltmeter does not disturb the circuit.
Read the worked answer
The voltmeter reads the potential difference across the 3 Ω resistor: $V = IR = 1.09 \times 3 \approx 3.27$ V.
Check with the full loop: the 6 Ω drops $1.09\times6 = 6.55$ V, the 3 Ω drops 3.27 V, the ammeter drops $1.09\times2 = 2.18$ V; sum $= 6.55 + 3.27 + 2.18 = 12.0$ V. ✓
An ideal voltmeter has infinite resistance, so it draws no current and does not change the series current. A real voltmeter would have finite resistance, placing it in parallel with the 3 Ω, lowering the equivalent resistance of that section and slightly altering the current - which is why ideal-voltmeter assumptions are stated explicitly.