Shorting a Bulb: What Happens When You Bypass One
1 · Predict
Two bulbs (lampA, lampB) are in series. A switch, wired in parallel across lampB, is initially open. If you close it — creating a zero-resistance shortcut around lampB — what happens to lampA's brightness?
- Closing the switch bypasses lampB, so lampA now gets the full circuit current — it brightens.
- LampA dims when the switch closes.
- LampA's brightness doesn't change.
2 · Set Up
- Open the short-a-bulb preset and press Reset. Both bulbs start at equal resistance (100 Ω); the switch starts open.
- Enable the current readout on both bulbs.
- Record lampA's current with the switch open, then close the switch and record it again.
3 · Collect Data
| Switch state (0 = open, 1 = closed) | LampA current I (A) |
|---|---|
| 0 | |
| 1 |
Make a bar chart comparing lampA's current in the open-switch and closed-switch cases.
4 · Analyze
- With the switch open, both bulbs are in series: I = E/(2R) using E = 9 V, R = 100 Ω. With it closed, lampB (100 Ω) is bypassed by the 0 Ω switch, leaving just lampA: I = E/R.
- Explain why a zero-resistance path in parallel with lampB carries all the current, leaving zero current through lampB itself once the switch closes.
5 · Extend
- LampA's power roughly doubles when the switch closes (same reasoning as switched-lamp's I²R). Where does that extra power come from — was the battery working equally hard in both cases?
- A real switch has a tiny but nonzero resistance when closed. Would that change the qualitative result here (lampA brightening), or only the exact numbers?
The Physics Behind This Experiment
Shorting a Branch
A zero-resistance path in parallel with a component carries the current instead of that component (since current follows the path of least resistance), effectively removing it from the circuit — this is called 'shorting out' the component.