0
\$\begingroup\$

If I have this circuit in which the switch opens at t = 0 (after being closed for an infinite time, making the circuit steady state), for which I'm trying to find the voltage across R2 as a function of time, I need to determine the current through the inductor as a function of time first. To do this, I need to find the initial current through the inductor. (ignore element values)

schematic

simulate this circuit – Schematic created using CircuitLab

I can replace the inductor with a short circuit and use circuit simplifications to reduce the circuit to the following. This is before the switch is opened at t = 0, and use the current through this circuit as the initial current through the inductor.

schematic

simulate this circuit

After t = 0, my gut tells me I can simplify the circuit into two parts, one containing the inductor, and the other without.

schematic

simulate this circuit

What law states I can do this? Is it because opening the switch means there are no current loops between either side of the circuit?

Thanks for your help!

\$\endgroup\$
  • 2
    \$\begingroup\$ Could you shrink the size of your second and third diagrams? They make the question hard to follow. Also, please clarify whether you are saying that the second and third diagrams are before switch closure, or after? \$\endgroup\$ – gwideman Apr 8 '14 at 0:43
  • \$\begingroup\$ Also, is there supposed to be some correspondence between the resistor names in the first diagram, and those in the second and third? To be honest, I'm not seeing much valid reasoning I can latch onto here. \$\endgroup\$ – gwideman Apr 8 '14 at 0:49
0
\$\begingroup\$

This is resolved through Ohm's law. If you think of the switch as an infinite resistance, then if any current flows on the wire directly above the switch in the schematic, then an equal amount in the opposite direction must flow in the infinite resistor (switch) because there's no other path for the current to come back from. Any current at all in an infinite resistor implies infinite voltage. You can assume that you never have infinite voltage in any realistic system. Therefore, the current through the infinite resistor must be 0 which implies that the current through the wire above it must also be 0.

Also, a circuit implies closed loop as seen here: "An electrical circuit is a network consisting of a closed loop, giving a return path for the current." http://en.wikipedia.org/wiki/Electrical_circuit

Perhaps I should also mention that Kirkchhoff's Voltage Law would also get interesting if you had current going through the switch. That would imply infinite voltage which would then imply negative infinite voltage across the rest of the circuit going through the top wire due to KVL. Obviously, it's not possible.

The only caveat that I should mention here is something like an antenna where there physically doesn't have to be a closed loop return path. This is resolved through capacitance to ground and the fact that one uses a high frequency wave to send short pulses through the inherent capacitance to "ground". The point I'm making is that through capacitance in a circuit, you can send current through an "open-loop", but that can only be done for very short periods of time. That's a different area of study than what you're doing here. You're doing pulsed-dc analysis whereas AC analysis can get interesting.

\$\endgroup\$
  • \$\begingroup\$ Hey, thanks, all of what you said makes a lot of sense. I really appreciated your multiple approaches. \$\endgroup\$ – Matt Egan Apr 8 '14 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.