The Overheating Battery Pack
A student is building a portable reading lamp for a design project. The lamp uses a 9 V battery, a 3 W LED module rated at 3.2 V, a resistor, and a push-to-make switch. On the first test the LED lights brightly, but after about four minutes the resistor becomes too hot to touch and the plastic battery clip begins to soften. The student measured the current at 0.35 A and the battery voltage at 8.7 V under load. The resistor is a ¼ W carbon film type marked 22 Ω. The lamp housing is a sealed 3D-printed box with no ventilation holes.
1. Calculate the power dissipated in the resistor and explain why it is failing.
Read the worked answer
Using P = I²R with the measured current:
P = 0.35² × 22 = 0.1225 × 22 ≈ 2.7 W
Check with the voltage across the resistor: the LED drops about 3.2 V, so the resistor sees roughly 8.7 − 3.2 = 5.5 V. Then P = V²/R = 5.5²/22 ≈ 1.4 W.
The two figures differ because the measured current and the assumed LED drop are not consistent; the true operating point is somewhere between them. Either way, the resistor is dissipating well over its ¼ W (0.25 W) rating - between about 5 and 11 times its rating. A carbon film resistor run far above its rating heats rapidly, and in a sealed box with no ventilation the heat cannot escape, so the plastic clip softens. The resistor is not faulty; it is simply the wrong component for the job.
2. Propose a revised circuit that runs the LED within its ratings and explain the reasoning behind each value.
Read the worked answer
Work from the LED's needs. A 3 W LED at 3.2 V draws about 3 / 3.2 ≈ 0.94 A, which is far more than a 9 V battery can supply comfortably. A better approach is to use a lower-power LED or a proper driver.
Option A - keep the 9 V battery, use a 1 W LED. A 1 W LED at 3.2 V draws about 0.31 A. The resistor must drop 9 − 3.2 = 5.8 V at 0.31 A, so R = 5.8 / 0.31 ≈ 19 Ω (use 18 Ω or 22 Ω). Power in the resistor = 0.31² × 18 ≈ 1.7 W, so specify a 3 W or 5 W resistor, not a ¼ W part. This wastes a lot of energy as heat.
Option B - use a constant-current driver or a 3.7 V lithium cell with a boost driver. This is more efficient and avoids the hot resistor entirely.
In both cases, add ventilation slots to the housing and keep the resistor away from the plastic clip. The key reasoning: choose the resistor from the required current and voltage drop, then choose its power rating at least double the calculated dissipation.