Two Bulbs, One Supply: Diagnosing a Dim Lamp
A student builds a series circuit with a 12 V battery of negligible internal resistance, a 6 Ω resistor, and a small lamp rated 6 V, 3 W. The lamp glows dimly. The student's write-up states: "The lamp is dim because the current is used up as it goes around the circuit, and the 6 Ω resistor takes most of it." A second student measures the current with an ammeter and gets 0.50 A, then claims the lamp is receiving only 3 W and should therefore be at full brightness. The teacher needs to address both the conceptual error and the quantitative claim without simply giving the answer.
1. Calculate the lamp's rated current and resistance, then determine the actual current and power in this circuit. Evaluate the second student's claim.
Read the worked answer
Lamp ratings
Rated current: I = P/V = 3/6 = 0.50 A. Rated resistance: R = V²/P = 36/3 = 6 Ω (assumed constant, though a real filament's resistance rises with temperature - worth flagging).
Actual circuit
Total resistance = 6 Ω (resistor) + 6 Ω (lamp) = 12 Ω.
I = 12/12 = 1.0 A.
Hmm - the student measured 0.50 A. That is a factor of two low, which is itself a red flag: either the resistor is not 6 Ω, the battery is not 12 V under load, or the ammeter is misranged/misread.
Taking the measured 0.50 A at face value:
- Total R = 12/0.50 = 24 Ω, so the lamp branch is ~18 Ω - far above its rated 6 Ω.
- Power in lamp = I²R = (0.50)²(18) = 4.5 W; or if R = 6 Ω, P = (0.50)²(6) = 1.5 W.
Evaluating the claim: the second student's error is arithmetic - 0.50 A through a 6 Ω lamp gives 1.5 W, not 3 W. Even at the rated 3 W the lamp is at its design power, not "full brightness" beyond it. And 1.5 W is a quarter of rated power, which looks distinctly dim - consistent with the observation.
Key point: the measured current contradicts the component values, so the circuit must be re-examined before any power conclusion is drawn.
2. Address the first student's "current is used up" statement. What model would you substitute, and what quick demonstration would confront the misconception?
Read the worked answer
The misconception: "current is used up" treats current as a consumable fluid. In a series loop, charge is conserved - the current is the same at every point. What is "used up" is energy, transferred from the charges to the components as they pass through.
Substitute model:
- Current = rate of charge flow, identical everywhere in a series loop (Kirchhoff's junction/continuity).
- Each component converts electrical energy to other forms; the potential difference across a component is the energy transferred per unit charge.
- The battery supplies energy per unit charge (its emf); the resistors and lamp dissipate it. Sum of p.d.s = emf.
So the lamp is dim not because current is depleted but because the p.d. across it is a fraction of the supply, set by the resistance ratio.
Demonstration: place two identical ammeters on opposite sides of the lamp in the series loop and show both read the same value. Then add a voltmeter across the resistor and across the lamp and show the p.d.s add to the supply voltage. This makes the energy-per-charge idea visible and directly contradicts "used up."
Follow-up question to pose: "If current were used up, what would the ammeter after the lamp read?" Let students commit to a prediction before measuring.