- #1
DemoniWaari
- 15
- 0
Hello there!
Now if I have a very very simple RL-circuit with a voltage source and we have a switch which is suddenly flipped in the off-position... does the SAME differential equation apply as if it's flipped ON?
This I mean:
[itex] \frac{\partial I }{\partial t}L + RI = U [/itex]
I do understand how that is formed when the switch is flipped ON, but if it's flipped OFF doesn't that mean that there goes NO current through, and thus that equation is kinda pointless?
Now my friend says that in his book it says that after it's flipped off the equation is:
[itex] \frac{\partial I }{\partial t}L = RI [/itex]
I don't get it where the [itex] U [/itex] suddenly vanishes and neither does she so I though I'd give it a shot here.
Now if I have a very very simple RL-circuit with a voltage source and we have a switch which is suddenly flipped in the off-position... does the SAME differential equation apply as if it's flipped ON?
This I mean:
[itex] \frac{\partial I }{\partial t}L + RI = U [/itex]
I do understand how that is formed when the switch is flipped ON, but if it's flipped OFF doesn't that mean that there goes NO current through, and thus that equation is kinda pointless?
Now my friend says that in his book it says that after it's flipped off the equation is:
[itex] \frac{\partial I }{\partial t}L = RI [/itex]
I don't get it where the [itex] U [/itex] suddenly vanishes and neither does she so I though I'd give it a shot here.