troubleshooting
Why are two LEDs in series so dim?
Short answer: Each LED takes about 2 V before it conducts. Two in series need 4 V, leaving a 4.5 V pack only 0.5 V to push current. Use 9 V, or wire them in parallel.
See it happen
This is the broken circuit, simulated live. The readout under the board tells you what the parts are doing.
simulating…
What's going on
In a series loop the same current flows through everything, and each part takes its share of the battery's voltage. Each red LED insists on about 2.0 V before it lets current through — that's its forward voltage, and it doesn't shrink because you added a second one.
- Two LEDs: 2.0 V + 2.0 V = 4.0 V spoken for
- 4.5 V − 4.0 V = 0.5 V left for the resistor
- 0.5 V ÷ 220 Ω = 2.3 mA — a glow, not a light
Real LEDs vary a little, and the simulator uses a proper diode curve rather than a flat 2 V, so you'll see a slightly different number on the board — same story.
The fix
Give the loop more voltage, then re-pick the resistor:
- 9 V battery: 9 − 4 = 5 V spare; 5 V ÷ 470 Ω ≈ 10.6 mA. Both LEDs bright.
- Or wire them in parallel on 4.5 V, each with its own 220 Ω. Each branch is a separate loop that sees the full 4.5 V.
Don't share one resistor between parallel LEDs — the one with the slightly lower forward voltage hogs the current.
simulating…
Also check
- Green LEDs are about 2.1 V and blue/white about 3.0 V: two blues in series need more than 6 V, so 3×AA can't light them at all.
- Three or more red LEDs in series want a 9 V battery even for a dim glow.
- Mixed colours in series all carry the same current but won't look equally bright — that's the LEDs, not a fault.