Two Bulbs, One Cell, and "Used-Up Current"
Students build a series circuit with one 1.5 V D cell and two identical bulbs. With the bulbs glowing, multimeter readings are 1.48 V across the cell, 0.72 V across bulb 1, and 0.71 V across bulb 2. When a student loosens bulb 1, bulb 2 goes dark. A second group wires the same two bulbs in parallel across the same cell; both bulbs are noticeably brighter, and the cell voltage under load reads 1.35 V. During the whole-class discussion, a student explains, "The first bulb uses up some of the current, so the second bulb gets less - that's why they're dimmer than the parallel ones." The teacher has two more identical bulbs, extra wire, and one multimeter, and must decide whether to address the claim now or move on to the planned lesson on Ohm's law.
1. Use the voltage readings to evaluate the "used-up current" explanation, and explain what the readings do support.
Read the worked answer
The readings contradict the student's mechanism. If current were consumed by bulb 1, bulb 2 would receive less current, and since the bulbs are identical, less current would mean a smaller voltage drop across bulb 2. Instead the drops are essentially equal - 0.72 V and 0.71 V - which is what a single current passing through two equal resistances produces: V = IR with the same I and the same R gives the same V. The two drops sum to 1.43 V, close to the 1.48 V terminal value, with the small remainder attributable to wire and meter leads. Loosening bulb 1 extinguishing bulb 2 confirms a single path, not a depleted supply at the far end. The parallel result is the telling comparison: there each bulb receives nearly the full cell voltage, so each is brighter, and the cell's terminal voltage sags to 1.35 V because the total current drawn is larger. Current is not used up; energy is transferred to light and heat.
2. Design a quick test with the available materials that would help students distinguish "current is used up" from "current is the same everywhere in a series loop," and state the predicted result of each model.
Read the worked answer
Insert the multimeter in series at three points: between the cell and bulb 1, between bulb 1 and bulb 2, and between bulb 2 and the cell. The "used-up" model predicts a measurable drop at each position, so the reading after bulb 1 would be noticeably smaller than the reading before it and would fall again after bulb 2. The single-path model predicts the same current at all three points, within meter tolerance. A typical result might be 0.31 A, 0.31 A, 0.31 A. A second, faster check reinforces it: swap the bulb positions. If current were consumed, the bulb nearest the cell would always be brighter; in fact the brightness is unchanged by swapping, which is a strong and easily remembered result. The teacher should run the swap first - it takes two minutes - and use the three-point measurement as the explanation, then connect both to V = IR so the lesson on Ohm's law is served rather than postponed.
3. A student asks why the parallel bulbs dim the cell's voltage to 1.35 V. How should the teacher explain this without introducing internal resistance as a formula?
Read the worked answer
Treat the cell as a source with its own small resistance and use the class's own numbers as the anchor. In series, the total resistance is large, so the current is small, and only a little voltage is "lost" inside the cell - the terminal reading stays near 1.48 V. In parallel, two paths mean roughly twice the current, and the same small internal resistance now removes a larger share of the cell's voltage, so the terminal reading falls to 1.35 V. The key idea to state explicitly: the 1.5 V printed on the label is the electromotive force, and the voltage you measure at the terminals depends on how much current you demand. A useful check is to have students predict what happens with three bulbs in parallel - lower terminal voltage still, and a warmer cell. This keeps the explanation qualitative and grounded in the measured values, and it gives a concrete reason the printed voltage and the measured voltage are not the same number.