# Transient Network Analysis

Discussion in 'Homework Help' started by dalam, Jan 3, 2015.

1. ### dalam Thread Starter Member

Aug 9, 2014
58
6

I found out the voltage Vc(t) = 80-40e^(-1000t).
Voltage at t=1 msec ; Vc=80-40/e =65.28 Volts.
Using phasor I can find out Vr=√(80^2-65.28^2)=46.23 Volts.
Now I use P=Vr^2/R to find the power dissipation in resistor.
I end up getting P=21.378Watts. But this is not correct answer.
What am I doing wrong? Answer is 2.1Watts.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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It is not obvious to me what is meant by the function u(t)?

Aug 9, 2014
58
6
4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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Ok. Here is what I get: V(R) = V(P) because the red trace falls perfectly on top of the green trace. V(p) is the correct equation for the voltage across the resistor.

The blue trace is the power in the resistor.

5. ### MrAl Distinguished Member

Jun 17, 2014
2,551
515
Hi,

It's not that the u(t) is unusual, it's the whole 40+40 u(t) statement for the source. This implies that there was 40 volts present before t=0, and after t=0 there is another 40 volts present because of the step 40 u(t). The power at t=0.001 seconds therefore comes out to 2.1 watts, but that's not exact, more exact would be 2.17 watts but that's not exact either, so if you have a more exact result you can post that and we can compare (i'll post the much more exact result).

So your time expression is wrong to begin with. Perhaps you can show how you approached this problem.

6. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
That is what I concluded, too...

7. ### MrAl Distinguished Member

Jun 17, 2014
2,551
515
Hi Mike,

Wow we posted at the exact same time previously

But anyway, the source originally 40+40 u(t) could be interpreted as a simpler source of 80 u(t) and apply the t<0 existing voltage as the capacitor initial voltage. This leads to a single term equation that is an exponential.

8. ### dalam Thread Starter Member

Aug 9, 2014
58
6
I used equation Vc(t)=[v(0+)-v(∞)]*e^(-t/Γ)+v(∞)
Vc(0+)=40V
Vc(∞)=80V
Γ(time constant)=10^-3 sec.

9. ### dalam Thread Starter Member

Aug 9, 2014
58
6
Thank you for the simulation. But will you please elaborate on what am I missing in my solution?

10. ### ericgibbs AAC Fanatic!

Jan 29, 2010
2,588
389
hi,
Your mistake, is that the source voltage was at 40V at t<0, so after t>= 0, only 40 u[t] was applied across the series res/cap network.
ie: at t<0 the the circuit had a existing steady state voltage of 40V.

E

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11. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Your expression for voltage across the resistor vs time was incorrect. I was showing you the correct one V = 40e^(-t/R1C1) at the behavioral voltage source B1, which produces the same voltage at V(p) that the actual circuit produced at V(r).

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12. ### dalam Thread Starter Member

Aug 9, 2014
58
6
Thanks a lot. I got the correct answer now.
Vr(1ms)=(40*e^-1)^2/100= 2.165 Watts

13. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I don't understand where phasors are coming into play at all. Phasors are for sinusoidal signals, which this is not.

14. ### dalam Thread Starter Member

Aug 9, 2014
58
6
Yes, you are correct. I was doing network analysis after some time so it got messed up.

15. ### MrAl Distinguished Member

Jun 17, 2014
2,551
515
Hi,

Your result looks much better now.