The Battery That Would Not Charge
A student builds a simple circuit with a 9 V battery, a 100 Ω resistor, and an LED rated for 2.0 V and 20 mA. When connected, the LED is dim and the resistor becomes warm. The student measures 9.0 V across the battery, 7.0 V across the resistor, and 2.0 V across the LED. A classmate says the battery must be dead because the LED is dim. The student wants to know whether the circuit is wired correctly and how to make the LED brighter without exceeding its rating. A second identical LED is available, along with resistors of 50 Ω, 220 Ω, and 470 Ω.
1. Use the voltage measurements to judge whether the circuit obeys circuit rules and whether the battery is dead.
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The measurements are consistent with a working battery and a correct series circuit. In a series circuit the source voltage equals the sum of the drops: 7.0 V + 2.0 V = 9.0 V, which matches the battery. The current is I = V/R = 7.0 V / 100 Ω = 0.070 A, or 70 mA, which is far above the LED's 20 mA rating. That explains the warmth in the resistor and warns that the LED is being overdriven, not underdriven. The dim appearance may reflect the LED's behavior or a poor connection, but the numbers do not support a dead battery. The classmate's inference is wrong because dimness was attributed to the source without checking the current. A correct answer computes the current and flags the overcurrent as the real problem.
2. Recommend a change that makes the LED operate near its rated 20 mA and explain the calculation.
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To get about 20 mA with a 2.0 V LED on a 9 V source, the resistor must drop 7.0 V at 0.020 A, so R = 7.0 / 0.020 = 350 Ω. Among the available choices, 470 Ω is the safe pick because 50 Ω and 220 Ω would allow more current than rated. With 470 Ω the current is 7.0 / 470 ≈ 0.015 A, or 15 mA, which is below the 20 mA rating and safer for the LED. If more brightness is needed, the student could use the 220 Ω resistor only if the resulting current, about 32 mA, is acceptable for the specific LED, which the rating says it is not. The defensible choice is 470 Ω, accepting slightly lower brightness to stay within the device rating, and the student should verify the actual forward voltage rather than assume 2.0 V exactly.