Discussion in 'Homework Help' started by Thinker.93, Mar 22, 2013.

1. ### Thinker.93 Thread Starter New Member

Mar 22, 2013
1
0
Hi,

when we have a circuit that consists of a resistance and a capacitor and an inductor with a switch that is open at t<0 and closed at t=0
(ALL THE ELEMENTS OF THE CIRCUIT ARE CONNECTED IN SERIES),
what should i do with the capacitor and inductor ?!

should i solve this problem as the same way i solve circuits that is closed at t<0 ? thus, put a S.C instead of the inductor and an O.C instead of the capacitor ?

I mean,
do we replace these things only when there is a DC current ? OR we replace them in every t<0 even if there is no DC current??
I WANT TO KNOW THE RULE

finally, if we don't replace them, so what will i do ?

Last edited: Mar 22, 2013
2. ### WBahn Moderator

Mar 31, 2012
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7,654
More importantly, you need to understand WHY you ever replace them at all! If you understand that, then you can answer questions like this based on that understanding.

When a circuit reaches DC steady state, meaning that whatever voltages and currents are present are steady and unchanging, then a capacitor will not have any current flowing in it because, if it did, the voltage would be changing and the circuit would not be in steady state. But it can have a voltage across it, just as long as that voltage isn't changing. Similarly, an inductor can have a current in it but that current must not be changing. Thus, it cannot have any voltage across it because, if it did, the current would be changing.

So a capacitor looks like a "device" that can have any voltage across it but no current. That's a good description of an open circuit. Inductors, on the other hand, look like a "device" that can have any current flowing through it but no voltage across it. That's a good description of a short circuit.

In order to make these substitutions, the circuit HAS to be in DC steady state. So when you have a prblem with a switch changing position at some moment in time and you aren't given further details, you assume that the switch has been in its initial position long enough for the circuit to reach DC steady state. Once the switch it open, you can't make that assumption because it is a pretty safe bet that it is NOT in DC steady state! The change in the switch position has changed the conditions and now the circuit must respond and the voltages and currents will change as a result.

But you also know that, due to conservation of energy, the voltage across a capacitor and the current through an inductor cannot change instantaneously. Hence whatever those quantities were just prior to the change in the switch will be what they are ust after the change, though they will immediately start changing from there.