Two Resistors, One Battery: Reading a Circuit Correctly
A student builds a circuit with a 12.0 V ideal battery, a 6.0 Ω resistor R1 in series with a parallel combination of a 12.0 Ω resistor R2 and a 4.0 Ω resistor R3. An ammeter (ideal) reads the total current from the battery, and a voltmeter (ideal) is connected across R2. The student predicts the voltmeter will read 12.0 V because 'the parallel part gets the full battery voltage'. During the experiment the 4.0 Ω resistor heats up noticeably, and its resistance rises to 6.0 Ω. The battery remains at 12.0 V.
1. Compute the ammeter and voltmeter readings before heating, and judge the student's prediction.
Read the worked answer
Step 1 - Parallel equivalent. R2 ∥ R3: (12 × 4)/(12 + 4) = 48/16 = 3.0 Ω.
Step 2 - Total resistance. R_total = 6.0 + 3.0 = 9.0 Ω.
Step 3 - Total current. I = 12.0 / 9.0 = 1.33 A (ammeter).
Step 4 - Voltage across the parallel block. V_p = I × R_p = 1.33 × 3.0 = 4.0 V (voltmeter).
Judgement: The student is wrong. R1 drops the remaining 8.0 V. Only the parallel block sees 4.0 V, not 12.0 V, because R1 is in series and shares the battery voltage.
2. After the 4.0 Ω resistor heats to 6.0 Ω, recompute both readings and state the direction of change.
Read the worked answer
Step 1 - New parallel equivalent. 12 ∥ 6: (12 × 6)/(12 + 6) = 72/18 = 4.0 Ω.
Step 2 - New total resistance. 6.0 + 4.0 = 10.0 Ω.
Step 3 - New current. I = 12.0 / 10.0 = 1.20 A. The ammeter reading falls from 1.33 A to 1.20 A.
Step 4 - New parallel voltage. V_p = 1.20 × 4.0 = 4.8 V. The voltmeter reading rises from 4.0 V to 4.8 V.
Judgement: Increasing R3 raises the parallel equivalent, so it takes a larger share of the battery voltage while total current falls. Both changes are consistent with series voltage division.
3. Explain why the total current does not simply fall in proportion to the increase in R3, using the limiting cases.
Read the worked answer
Reasoning: R_total = R1 + (R2 R3)/(R2 + R3). As R3 → 0, the parallel term → 0 and R_total → R1 = 6.0 Ω, giving I → 2.0 A. As R3 → ∞, the parallel term → R2 = 12.0 Ω and R_total → 18.0 Ω, giving I → 0.67 A.
Judgement: The total current is bounded between 0.67 A and 2.0 A, so a change in R3 from 4 Ω to 6 Ω produces only a modest current change (1.33 A to 1.20 A). The parallel branch 'absorbs' part of the change. Recall of a formula is not enough; the student must reason about limits to see why the effect is small.