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

[eehelp150]

Homework Statement

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?

 

[gneill]

What do I do with the initial condition?

It's the integration constant.

 

[NascentOxygen]

$v_C(t)\ =\ V_C(0)\ +\ \frac 1C \displaystyle\int_{0}^{t} ...$

 

[eehelp150]

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...

 

[cnh1995]

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^-.

 

[eehelp150]

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)

 

[cnh1995]

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.

 

[eehelp150]

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?

 

[cnh1995]

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.

 

[eehelp150]

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.

 

[cnh1995]

10000sin(0+k+pi/4) = 3

This is not where you add the integration constant.

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?

 

[gneill]

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 $5000 \sqrt{2} - 3$.

 

[NascentOxygen]

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 V_C(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.

 

[NascentOxygen]

Your friend's approach would be appropriate had the problem specified:
iC(t) = 10cos(1000t + $\theta$[/color] )
Vc(0-) = 3v

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