A Battery, a Bulb, and a Mystery Voltage
Students build a circuit with a 6.0 V battery pack, a small bulb rated 6.0 V and 0.50 A, and a length of nichrome wire used as a variable resistor. With the wire at full length, the bulb glows dimly and a multimeter reads 4.2 V across the bulb. As they shorten the wire, brightness increases; at one setting the bulb reads 5.8 V and the wire becomes noticeably warm. A student connects the multimeter across the battery and reads 5.9 V at that setting, whereas at full wire length the battery read 6.0 V. Another student reports a 'current' reading of 0.05 A taken with the meter set to the 10 A range. The group must explain the readings and decide whether the bulb is operating near its rating.
1. Explain why the battery voltage changes with the wire setting and why the wire heats up. Use the readings to support the reasoning.
Read the worked answer
The battery has internal resistance. Terminal voltage equals emf minus the internal drop (V = emf - I·r). At full wire length, total resistance is high, current is small, and the terminal voltage is near the 6.0 V emf. When the wire is shortened, total resistance falls, current rises, and the internal drop grows, so terminal voltage falls to 5.9 V while the bulb voltage rises toward the source. The observed 6.0 V to 5.9 V shift is small but in the expected direction.
The nichrome wire heats because power dissipates in any resistor carrying current (P = I²R). Shortening the wire lowers its resistance, which raises current; the remaining wire carries more current and warms noticeably. The bulb brightens because the voltage across it rises from 4.2 V to 5.8 V, increasing its power. The wire is acting as a series control resistance, not as an ideal lossless element, which is why the circuit is not a simple fixed-voltage source driving the bulb.
2. Assess whether the bulb is near its 6.0 V, 0.50 A rating at the 5.8 V setting, and identify what additional measurement is needed.
Read the worked answer
At 5.8 V the bulb is close to its rated voltage, but the rating also specifies 0.50 A, and current was not validly measured. The reported 0.05 A on the 10 A range is suspect: a 10 A range typically resolves only to about 0.01 A, so a reading of 0.05 A has roughly 20% resolution uncertainty at best, and a small current on a high range is unreliable. The reading may also reflect the meter's own burden resistance altering the circuit.
To decide, measure current on a lower range (for example 200 mA or 2 A) in series with the bulb, or compute it from the bulb's resistance at operating temperature once the bulb voltage is known. Note that bulb resistance rises with filament temperature, so a cold-resistance calculation will overestimate current. The defensible statement is that the bulb is near rated voltage and probably near rated current, but the current claim needs a proper measurement before it can be asserted.
3. A teammate says the bulb 'uses up' current, so current is lower after the bulb than before it. Correct this and explain what is actually conserved.
Read the worked answer
Current is not consumed by a bulb. In a series circuit the same current passes through every element at any instant; charge is conserved, and charge does not pile up or disappear in the bulb. What the bulb converts is energy: electrical energy becomes light and heat, which shows as a voltage drop across the bulb, not a current drop.
The confusion often comes from conflating current with energy. The correct accounting is that the battery supplies energy per unit charge (emf), and each element takes a share of the voltage drop proportional to its resistance in a series loop. If students measured different currents at two points in the series loop, that would indicate a parallel path or a measurement error, not consumption. This distinction matters for reading the earlier data: the bulb dimming at full wire length reflects reduced current everywhere in the loop, not current being drained before reaching the bulb.