Find Vc(t) given iC(t): Initial Condition Included

  • Thread starter eehelp150
  • Start date
In summary, the problem asks for the nett charge that will be added to a capacitor after a sinusoidal input current has passed through it. The answer is that the nett charge will be zero.
  • #1
eehelp150
237
0

Homework Statement


upload_2016-10-30_16-34-51.png


iC(t) = 10cos(1000t+pi/4)
Vc(0-) = 3v

Homework Equations


ic = C * d/dt*Vc(t)

The Attempt at a Solution


Vc(t) = 1/C * integral(Ic(t)dt) =
1/10^-6 * 10sin(1000t+pi/4)/1000 = 10000sin(1000t+pi/4)

What do I do with the initial condition?
 
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  • #2
eehelp150 said:
What do I do with the initial condition?
It's the integration constant.
 
  • #3
##v_C(t)\ =\ V_C(0)\ +\ \frac 1C \displaystyle\int_{0}^{t} ... ##
 
  • #4
gneill said:
It's the integration constant.
1/10^-6 * 10sin(1000t+pi/4)/1000 = 10000sin(1000t+pi/4) + K

Vc(0-) = 3
10000sin(0+pi/4) + K = 3
7071.0678 + K = 3
K = -7068.06781186547524400844362104
?
That doesn't seem right...
 
  • #5
eehelp150 said:
1/10^-6 * 10sin(1000t+pi/4)/1000 = 10000sin(1000t+pi/4) + K

Vc(0-) = 3
10000sin(0+pi/4) + K = 3
7071.0678 + K = 3
K = -7068.06781186547524400844362104
?
That doesn't seem right...
See NascentOxygen's post above. That's the formula you should use. Integration constant is nothing but Vc at t=0-.
 
  • #6
cnh1995 said:
See NascentOxygen's post above. That's the formula you should use. Integration constant is nothing but Vc at t=0-.
sin(1000t+pi/4)/100 [0 t]

sin(1000t+pi/4)/100 + sin(pi/4)/100
3 + 1000000(sin(1000t+pi/4)/100 + sqrt(2)*50)
3 + 10000sin(1000t+pi/4) + 50000000sqrt(2)
 
  • #7
eehelp150 said:
1/10^-6 * 10sin(1000t+pi/4)/1000 = 10000sin(1000t+pi/4) + K

Vc(0-) = 3
10000sin(0+pi/4) + K = 3
7071.0678 + K = 3
K = -7068.06781186547524400844362104
?
That doesn't seem right...
Well, actually this is right. I verified it using the above formula. I didn't look through your solution in detail earlier. But it looks correct to me now.
 
  • #8
cnh1995 said:
Well, actually this is right. I verified it using the above formula. I didn't look through your solution in detail earlier. But it looks correct to me now.
Is there any way to simplify this so that there's no ugly constant?
 
  • #9
eehelp150 said:
Is there any way to simplify this so that there's no ugly constant?
No. The constant can't be eliminated because of the initial condition. You can make it look better by writing it as
Vc(t)=10sin(wt+pi/4)-7.068 kV.
 
  • #10
cnh1995 said:
No. The constant can't be eliminated because of the initial condition. You can make it look better by writing it as
Vc(t)=10sin(wt+pi/4)-7.068 kV.
I asked my friend how he did it and he showed me this:
10000sin(1000t+pi/4)
Vc(0-) = 3
10000sin(0+k+pi/4) = 3
sin(k+pi/4) = 0.0003
k + pi/4 = sin^-1(0.0003)
k = (sin^-1(0.0003)-45degrees)/1000 = 0.045s
Vc(t) = 10^4*sin(1000t)V
Is this even remotely correct or did he just bs it?
The reason I'm asking if it can be simplified further is because so far, most of the questions have had nice numbers.
 
  • #11
eehelp150 said:
10000sin(0+k+pi/4) = 3
This is not where you add the integration constant.
eehelp150 said:
Vc(t) = 10^4*sin(1000t)V
This does not give Vc=3V at t=0.

Also, I don't understand what problem will having that constant in the equation cause. "Nice numbers" and "ugly constants" do not matter in an equation as long as it is mathematically and technically correct. Is it specifically asked in the question to remove the constant?
 
  • #12
eehelp150 said:
Is there any way to simplify this so that there's no ugly constant?
Well, if you leave your calculator and root two alone the ugly constant is just ##5000 \sqrt{2} - 3##.
 
  • #13
eehelp150 said:
I asked my friend how he did it and he showed me this:
10000sin(1000t+pi/4)
Vc(0-) = 3
10000sin(0+k+pi/4) = 3
sin(k+pi/4) = 0.0003
k + pi/4 = sin^-1(0.0003)
k = (sin^-1(0.0003)-45degrees)/1000 = 0.045s
Vc(t) = 10^4*sin(1000t)V
Is this even remotely correct or did he just bs it?
The reason I'm asking if it can be simplified further is because so far, most of the questions have had nice numbers.
That is a creative approach, but it's not correct.

Over one full cycle (or period) the nett charge that the sinusoidal input current will have added to the capacitor is ZERO. This means that after each cycle of current, the capacitor voltage must back to where it started, i.e., it will have returned to VC(0). Your answer must show this.

In contrast, if the capacitor voltage were (as your friend claims) a pure sinusoid with no DC offset, it would not show this periodic return to V(0) at times of t = n⋅T, where T is the period and n is an integer.
 
  • #14
Your friend's approach would be appropriate had the problem specified:
iC(t) = 10cos(1000t + ##\theta## )
Vc(0-) = 3v

and so you'd be required to determine the ##\theta## that gives VC(0) as 3v. But this not what you are told about iC(t). You are not told it has some unknown phase; on the contrary, you are told it has some definite angle.
 

1. What is Vc(t)?

Vc(t) is the voltage across a capacitor at a specific time t.

2. How is Vc(t) related to iC(t)?

Vc(t) and iC(t) are related through the equation Vc(t) = (1/C) * ∫iC(t)dt + Vc(0), where C is the capacitance and Vc(0) is the initial voltage across the capacitor.

3. How do you find Vc(t) given iC(t) and the initial condition?

To find Vc(t), you can use the equation Vc(t) = (1/C) * ∫iC(t)dt + Vc(0), where C is the capacitance and Vc(0) is the initial voltage across the capacitor. You will need to integrate the current function, iC(t), over the desired time interval.

4. What is the significance of the initial condition when finding Vc(t)?

The initial condition, Vc(0), is the voltage across the capacitor at time t = 0. It is important to include this in the equation for Vc(t), as it affects the overall value of Vc(t).

5. Can you provide an example of finding Vc(t) given iC(t) and the initial condition?

Yes, for example, if iC(t) = 2t and Vc(0) = 5V, and we want to find Vc(3), we can use the equation Vc(t) = (1/C) * ∫iC(t)dt + Vc(0). Substituting in the values, we get Vc(3) = (1/2) * ∫2tdt + 5 = t^2 + 5. Therefore, Vc(3) = 14V.

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