**Ask a monkey a physics question thread**

[MadMojoMonkey]

The wire acting as a capacitor would be effectively shorted at first and then increase in resistance as it's charging until it's fully charged and then it stops conducting. If it's DC you'd only get a short spike. If it's AC it would act as a conductor with the phase shifted by 90°.

Yes, but the "short spike" from the DC happens over the entire wave passing down the wires from the switch/battery, right?

So like... there's the "immediate" response due to the distance between the switch and the light bulb. But then that wave of voltage/current flow expands out in the wires at c_wire. Causing a wave of those capacitave coupled responses that arrive at the light bulb. Each little nanosecond of spreading current creates another pulse that is overlapped with all the others.

But then, eventually, all the current gets out to that furthest distance (1/2 light second away), and starts coursing toward the light bulb.
Does the "short spike" last 0.5 s? Is that an 0.5 s plateau that then decays, or an impulse spike that has decayed to 0 after 0.5 s?

And how does our picture of the circuit change as we round those furthest distant points in the circuit and the current wave is now heading back toward the bulb?

Like, in terms of current at the bulb.
Is it a spike, dip and rise to V-0?
Is it a quick spike that holds at V_0?
Is it a gradual climb to V_0?

Something else? The transient ripples of the switch's discontinuity in time manifesting over presumably many seconds?

I'm fascinated by this.

[oskar]

Good point, I think you're right, it would probably spike up in exactly m/c and then actually stay up for the entire second it takes for the circuit to close. One thing is giving me a headache: If you think of it as a series of capacitors, then the last one will conduct at 0,5s, but the current will not see our load until 1s after the switch is turned on, or just as the actual real world usable circuit would be completed. So that means our series of capacitors have been conducting for only half a second, but they have been creating a current at the load for a full second. What does that mean for what we would measure at the load for that first second?