# Question about transients

Discussion in 'Homework Help' started by Teknolog, Dec 16, 2013.

1. ### Teknolog Thread Starter Member

Sep 1, 2013
31
0
Here's a circuit where the switch has been closed for a long time, so that Vc(t) is fully charged. At t=0 the switch opens. R1=R2.

When doing the transient analysis for what happens when t>0:
I would assume the the current goes in a clockwise direction in the loop betwen the capacitor and the resistance R2. Then we can write a KVL equation for this as

-Vc(t) + VR2 = 0

(since Vc(t) = VR2)

Which eventually (with diff.eq) gives the solution Vc(t)=Vo/2*e^(t/RC)

However this solution is incorrect because that would be an increasing voltage. But if we instead assume that the current goes COUNTERCLOCKWISE between Vc and R2, the solution becomes correct, since then we have

Vc(t) + Vr2 = 0

Which (eventually) gives the correct solution Vc(t)=Vo/2*e^-(t/RC)

But why is this correct? Doesn't the current actually go clockwise? Why does the solution become correct if we assume it to go counterclockwise which is not true?

Someone please explain!

File size:
2.1 KB
Views:
34
2. ### Teknolog Thread Starter Member

Sep 1, 2013
31
0
Actually I think I just confused myself (again), as it just has to do with the reference direction of the current, not the (necessarily) actual direction.

If I am correct now, we set the reference direction counterclockwise only because the voltage is given in a way that the passive reference direction (+ to - for voltage drops) says it should go this direction through the capacitance. That will in turn give a passive reference direction also for the resistor. Then we get the equation Vc(t) + VR = 0.

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,388
497
http://hyperphysics.phy-astr.gsu.edu/hbase/math/deinhom.html#c3

4. ### WBahn Moderator

Mar 31, 2012
17,737
4,789
You basically have it figured out, but perhaps it would help to be a bit more explicit on things.

First, you said that the switch had been closed for a long time and that the capacitor was "fully charged". What does "fully charged" mean. Even if you mean that it has reached steady state and has zero current in it, that really isn't enough. When dealing with these kinds of problems, you want to clearly identify the initial conditions that carry across the switching event -- namely the voltages across all capacitors and the currents in all inductors.

In this case,

Vc(t=0+) = Vc(t=0-) = Vo*R2/(R1+R2)

As you've noted that R1=R2, this reduces to

Vc(t=0+) = Vo/2

Be explicit about this kind of stuff; don't bury it inside of later work.

Next, as you've basically realized, you can declare your loop current to go in either direction, but the constitutive equations for all of the components are defined in terms of the passive sign convention. Thus, when we say that

v = R * i

v = L * di/dt

i = C * dv/dt

These are ALL defined with the current going through the device from the positive terminal to the negative terminal. You are free to choose one or the other -- you can decide which direction through the device corresponds to positive current or you can decide which terminal of the device is the positive terminal. But once you have done one of those, the other is decided as well and you need to be consistent with it.

So let's see how this shakes out with this circuit. Let's choose the loop direction to be clockwise. We'll call the loop current i and the capacitor current i_c.

Our basic KVL equation summing voltage gains in the direction of the loop current is

v_c(t) - i(t)*R = 0

i_c(t) = C * dv_c(t)/dt

By the passive sign convention, i_c(t) = - i(t), therefore

i(t) = - C * dv_c(t)/dt

v_c(t) + RC * dv_c(t)/dt = 0

With the loop current counterclockwise, our basic KVL equation summing voltage gains in the direction of the loop current is

- v_c(t) - i(t)*R = 0

i_c(t) = C * dv_c(t)/dt

By the passive sign convention, i_c(t) = i(t), therefore

i(t) = C * dv_c(t)/dt

v_c(t) + RC * dv_c(t)/dt = 0

And you can see that we end up with the identical differential equation with either choice.