Circuits
Short a bulb
Close the switch to short out lamp B; lamp A then gets the full voltage and brightens.
Short a bulb — interactive Circuits simulation. Close the switch to short out lamp B; lamp A then gets the full voltage and brightens. Free browser-based virtual physics lab with live SI measurements and a guided tutorial.
Shorting a branch
Closing a switch across a bulb bypasses it with a near-zero resistance path.
An ideal short pulls the voltage across lamp B to ~0 while lamp A gets more of the supply voltage and brightens.
- Short ≈ 0 Ω bypass
- Current prefers the low-R path
Investigation brief
Plan the question before you open the lab
The brief mirrors the prerendered page: driving question, competing predictions, variable roles, governing laws, setup, analysis and extension prompts remain visible and in this order.
Driving question
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?
Predictions to weigh
- LampA dims when the switch closes.
- LampA's brightness doesn't change.
- Closing the switch bypasses lampB, so lampA now gets the full circuit current — it brightens.
Variable roles
What you set:
- Switch state (0 = open, 1 = closed)
What you measure:
- LampA current I (A)
How the investigation runs
- 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.
Governing equation
Shorting a Branch — V = I·R
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.
What the printable worksheet asks students to work out
- 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.
Where this shows up beyond the lab
- LampA's power goes up four times when the switch closes — the current doubles, and P = I²R goes as the square of it (the same reasoning as switched-lamp's). 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?
- AP Physics 2 — Unit 11: Electric Circuits
- AP Physics C: Electricity and Magnetism — Unit 11: Electric Circuits
- IB Physics — B.5 Current and circuits
- General High School Physics — Electricity & DC circuits
- NGSS High School Physics — Electric current and magnetic fields
- Welcome to Short A Bulb
- Select the bulb A
- Press Play
- Bypass a bulb
- Open the Properties panel
- You did it!
Open the interactive simulation to build the scene, press Play, and explore with live measurements and a guided tutorial.