# Transient state - calculating initial conditions

Discussion in 'Homework Help' started by tomson, Sep 7, 2010.

1. ### tomson Thread Starter New Member

Sep 6, 2010
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0
I'm currently trying to pass exam from transient states in electrical circuits. Problem is that when solving problems I always struggle to set the initial conditions and steady state correctly. During my studies I've learned Kirchoff Law's, Ohm's Law but none of them is of any help when determining the initial values of voltage drop on capacitor or current of inductor. They're just helpful when calculating natural response. During my studies nobody taught me when and why voltage drop occurs and how it is related to elements of circuit, and other basic stuff like this. So I'm looking for a source to get this knowledge from.

For example on what base, Uc and iL are calculated here ? :
1.
Data : E = 10, R=5, L=1, C=1
Circuit :

Solution: for t(0-) it is obvious. Both Uc and iL are 0. At t(0+) Uc = 5, iL = 1 , but which resistor causes the drop on capacitor ? One from the first or second branch ? Or both ? iL is I believe just E/2R .

2.
Data : E=10, R=5, L=1H, C=100uF
Circuit:

Solution: for t(0-) iL = 2, Uc = 10. Why here is no voltage drop ? Only resistors before point of measurement influence voltage on capacitor ?

3.
Data: L=1, C=1, E=10, R=1
Circuit:

Solution: for t(0-) Uc=0, iL = 10. Why voltage on capacitor is zero ? Shouldn't it be equal to E ?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
In the first example I am assuming the switch closes at t=0+. Is that correct?

I would have thought that at t=0+, Uc=0 and iL =0. While at t=∞, Uc=5V and iL=1A.

In the second and third examples I am assuming the circuits have reached a steady state condition before t=0-. In either case the currents and voltages are then assumed to be at a steady-state (constant DC) condition.

If the current in an inductor is at steady-state (constant DC) then there can be no voltage drop across the inductor [since e=L(di/dt) and di/dt=0]. If the voltage across a capacitor is at steady state (constant DC) then there can be no current flow in the capacitor [since i=C(dv/dt) and dv/dt=0].

So in case 2 the inductor has no voltage drop and the capacitor must be supporting the full supply voltage (10V). The resistor in parallel with the capacitor is also supporting 10V so it must be carrying 2A. The fully charged capacitor will not be conducting current since a capacitor at a steady-state voltage does not conduct current. The current flowing into R from the source must then be flowing through the inductor. Hence iL=2A.

In the third case the inductor current is at steady state so there can be no inductor voltage. The capacitor is in parallel with the inductor so it must have the same voltage drop as the inductor - i.e. 0V. The resistor must be supporting the full source voltage so its current will be 10A. The capacitor (at constant 0V DC) will have no current flow so the 10A current in the resistor must also flow in the inductor on its return to the source.

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3. ### tomson Thread Starter New Member

Sep 6, 2010
16
0
This analytical way of solving problems is what I'd like to learn. But can you propose any good materials concerning this matter ? In books I've read there's usually some introduction to circuits theory, what is resistor, capacitor, sources, then KVL, KCL, Ohm's Law but no problems similar to this one from which you can learn what's really happening in the circuit.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The excellent reference on transient circuit analysis I use is simply entitled "Linear Circuits" by Ronald E. Scott. My edition is 1964 so it may be hard to obtain.

An old "rule of thumb" for inductors and capacitors under steady-state DC conditions is to view the inductor as a short circuit and the capacitor as an open circuit. Like any rule of thumb, take care when applying it and have a clear idea of what is really happening.

Last edited: Sep 7, 2010