Emergency Lantern Current and Battery Sag
A trainee is checking a 9 V battery-powered emergency lantern built from three white LEDs wired in parallel, each with its own 330 Ω series resistor. With a fresh alkaline battery, the open-circuit voltage is 8.95 V. When the lantern is switched on, the battery terminal voltage falls to 7.10 V and the total current measured at the battery positive terminal is 18 mA. The LEDs look dim, and the trainee's colleague says the 330 Ω resistors are too large and wants to replace them with 100 Ω. The trainee has a digital multimeter, a variable bench supply, and a spare 9 V battery. The lantern must run for at least 6 hours on a 9 V alkaline battery rated about 550 mAh. The LEDs have a forward voltage of about 3.0 V at 20 mA.
1. Using the bench supply set to 9.0 V, what current should each LED branch draw, and what total current should the trainee expect? Compare this with the 18 mA measured from the battery and explain the discrepancy quantitatively.
Read the worked answer
Each branch sees 9.0 V − 3.0 V = 6.0 V across 330 Ω, so I_branch = 6.0/330 = 18.2 mA. Three parallel branches should draw 3 × 18.2 = 54.5 mA total.
The battery measurement is only 18 mA, about one third of that. While delivering 18 mA, the terminal voltage fell from 8.95 V to 7.10 V, a drop of 1.85 V. That implies an equivalent series resistance of 1.85 V / 0.018 A ≈ 103 Ω - far too high for a fresh 9 V alkaline cell. So the fault is not the 330 Ω resistors; it is a high-resistance source path: dirty battery clip, poor breadboard/contact, exhausted cell, or a meter/range issue. Confirm by repeating on the bench supply at 9.0 V and measuring at the battery clip and at the LED common node to see where the drop occurs.
2. The colleague proposes replacing each 330 Ω resistor with 100 Ω. Evaluate this proposal against the LED rating and the 6-hour runtime requirement. Show calculations and state your recommendation.
Read the worked answer
100 Ω would give (9.0 − 3.0)/100 = 60 mA per LED, or 180 mA total for three branches. That is three times the 20 mA LED rating, so the LEDs would be overdriven and could degrade or fail. Runtime from a 550 mAh battery at 180 mA is 550/180 ≈ 3.1 h, below the 6 h requirement.
With the existing 330 Ω, the ideal total is 54.5 mA, giving 550/54.5 ≈ 10 h, which comfortably meets the requirement once the battery/contact fault is fixed. Recommendation: keep 330 Ω, clean or replace the battery clip and holder, and verify current with the bench supply. If more light is needed, use a more efficient LED/driver rather than reducing the resistors.