The Dimming Bulb in a Series Circuit
Students build a circuit with a 12 V battery of negligible internal resistance, a 6 Ω resistor, and a small lamp rated 12 V, 24 W, all in series. When switched on, the lamp glows dimly. A student measures 8.0 V across the lamp and 4.0 V across the resistor. Another student insists the lamp must be faulty because it is rated for 12 V. A third student proposes adding a second identical lamp in parallel with the first to make the circuit brighter. The group must decide what is actually happening and whether the parallel proposal helps.
1. Use the measured voltages and the lamp rating to determine the lamp's actual resistance and power, and judge whether the lamp is faulty.
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The lamp's rated values give R = V²/P = 144/24 = 6 Ω at full brightness. In the circuit the voltages sum to 12 V (8 + 4), consistent with a series loop. The current is set by the total resistance: I = 4.0/6 ≈ 0.67 A. The lamp's actual resistance is then 8.0/0.67 ≈ 12 Ω, higher than the rated 6 Ω. This is expected because a filament's resistance rises with temperature. The actual power is V × I = 8.0 × 0.67 ≈ 5.3 W, far below 24 W, which explains the dim glow. The lamp is not faulty; it simply is not receiving its rated voltage because the series resistor takes a share. The judgement is that the circuit design, not the lamp, causes the dimness.
2. Evaluate the proposal to add a second identical lamp in parallel with the first.
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Adding a lamp in parallel lowers the resistance of that section. Two 12 Ω lamps in parallel give 6 Ω, so the total circuit resistance falls from 18 Ω to 12 Ω and the current rises to about 1.0 A. The current through each lamp is about 0.5 A, less than before (0.67 A), so each individual lamp is dimmer, though the pair together draws more power. The resistor now takes 6 V, leaving 6 V across the lamp section, so each lamp runs below its previous 8 V. The proposal therefore does not make the original lamp brighter; it makes it dimmer and adds a second dim lamp. A better fix is to remove or reduce the series resistor, or use a supply matched to the lamp.