Intro to Power Systems: AC RLC circuit voltage calculations

[SumDood_]

Homework Statement

I am getting the wrong answer for Part B.
I don't understand if what I am doing for part C is correct.

Relevant Equations

Voltage divider rule

$V_{S} = 120V$
$f = 60Hz$
$L = 20mH$
$R = 10\Omega$

(a) Assume the capacitance is omitted. What is the resistor voltage?
$X_{L} = j\omega L = j2.4\pi$
$V_{R} = Vs\cdot \frac{R}{R+X_{L}} = 76.5067-j57.6847$
$V_{L} = Vs\cdot \frac{X_{L}}{R+X_{L}} = 43.493+j57.6847$

(b) What value of capacitance is required to make the resistor voltage have the same magnitude as the source voltage?
$$V_{R} = Vs\cdot \frac{R//X_{C}}{R//X_{C}+X_{L}}$$
$$R//C = \frac{R\cdot X_{C}}{R+X_{C}} = \frac{R\cdot \frac{1}{j\omega C}}{R+\frac{1}{j\omega C}} = \frac{R}{j\omega RC + 1}$$
$$V_{R} = Vs\cdot \frac{\frac{R}{j\omega RC + 1}}{\frac{R}{j\omega RC + 1}+j\omega L} = Vs\cdot \frac{R}{R+(j\omega L)(j\omega RC+1)}$$
$$V_{S} = V_{R} \equiv \frac{V_{R}}{V_{S}} = 1$$
$$1= \frac{R}{R+(j\omega L)(j\omega RC+1)}$$
$$(j\omega L)(j\omega RC+1) = 0$$
$$j\omega RC+1 = 0$$
$$C = \frac{-1}{j\omega R} = \frac{1}{120\pi\cdot 10} = 265.258\mu F$$

(c) What value of capacitance maximizes the resistor voltage? What is that maximum voltage?
$$V_{O} = V_{R} = Vs\cdot \frac{R}{R+(j\omega L)(j\omega RC+1)}$$
To increase the output voltage, I need to reduce the value of the denominator. I can do this by $(j\omega RC+1) = 0$ but this brings me back to the previous part of the question, which is incorrect for me.

 

[gneill]

I think you want to ponder what $\left| \frac{V_{R}}{V_{S}}\right|$ actually is. It's magnitude is one, but that means you need to take the magnitude of the expression on the right hand side.

$$V_{R} = Vs\cdot \frac{\frac{R}{j\omega RC + 1}}{\frac{R}{j\omega RC + 1}+j\omega L} = Vs\cdot \frac{R}{R+(j\omega L)(j\omega RC+1)}$$
$$V_{S} = V_{R} \equiv \frac{V_{R}}{V_{S}} = 1$$

You should get:
$$1=\left| \frac{R}{R+(j\omega L)(j\omega RC+1)}\right|$$

Then solve for C.

Edit: Hint: you should get two values for C.

 

[SumDood_]

$$1=\left| \frac{R}{R+(j\omega L)(j\omega RC+1)}\right|$$
I think I get it:
$$(j\omega L)(j\omega RC+1) = -2R$$
OR
$$(j\omega L)(j\omega RC+1) = 0$$

\begin{align*}
(j\omega L)(j\omega RC+1) &= -2R \\
-\omega ^{2}RLC + j\omega L &= -2R \\
-\omega ^{2}RLC &= -2R - j\omega L \\
C &= \frac{-2R - j\omega L}{-\omega ^{2}RL} = \frac{-20 - (j120\pi \cdot 20\cdot 10^{-3})}{-(120\pi)^{2}\cdot 10 \cdot 20\cdot 10^{-3}} = 751.96\mu F\angle 20.656^{\circ} \\ \\(j\omega L)(j\omega RC+1) &= 0 \\
(j\omega RC+1) &= 0 \\
C &= \frac{-1}{j\omega R} = \frac{-1}{j120\pi \cdot 10} = 265.258\mu F\angle 90^{\circ}

\end{align*}

I got the two values for C, but they are both wrong. Can you please tell me what I am doing wrong? I suspect I am making a mistake in the math, but I can't figure it out. Maybe I am not substituting correctly?

 

[The Electrician]

$$1=\left| \frac{R}{R+(j\omega L)(j\omega RC+1)}\right|$$
I think I get it:
$$(j\omega L)(j\omega RC+1) = -2R$$
OR
$$(j\omega L)(j\omega RC+1) = 0$$

\begin{align*}
(j\omega L)(j\omega RC+1) &= -2R \\
-\omega ^{2}RLC + j\omega L &= -2R \\
-\omega ^{2}RLC &= -2R - j\omega L \\
C &= \frac{-2R - j\omega L}{-\omega ^{2}RL} = \frac{-20 - (j120\pi \cdot 20\cdot 10^{-3})}{-(120\pi)^{2}\cdot 10 \cdot 20\cdot 10^{-3}} = 751.96\mu F\angle 20.656^{\circ} \\ \\(j\omega L)(j\omega RC+1) &= 0 \\
(j\omega RC+1) &= 0 \\
C &= \frac{-1}{j\omega R} = \frac{-1}{j120\pi \cdot 10} = 265.258\mu F\angle 90^{\circ}

\end{align*}

I got the two values for C, but they are both wrong. Can you please tell me what I am doing wrong? I suspect I am making a mistake in the math, but I can't figure it out. Maybe I am not substituting correctly?

The first expression in this post shows a fraction embedded between absolute value bars. You must rationalize that fraction. Multiply out the two expressions in parentheses in the denominator, them multiply the numerator and denominator by the complex conjugate of the denominator. This gets rid of "j" in the denominator. Do it like this:

When this is done you can separate the new numerator into a real part and imaginary part so it will look like a+jb. The you can take the square root of the sum of the squares of both parts.

You'll get sqrt(a^2 + b^2) This will be the magnitude of the expression you started with and it won't contain any imaginary parts. You can set it equal to 1 and solve.

 

[SumDood_]

@The Electrician
So I tried rationalization, but ended up with an expression that has j in the denominator:
\begin{align*}
\frac{R}{R+(j\omega L)(j\omega RC+1)} \cdot \frac{R-(j\omega L)(j\omega RC+1)}{R-(j\omega L)(j\omega RC+1)} &= \frac{R^{2}-R(-\omega ^{2}RLC + j\omega L)}{R^{2}-[(-\omega L)(-\omega ^{2}R^{2}C^{2}+j2\omega RC+1)]} \\ \\
&= \frac{R^{2}+\omega ^{2}R^{2}LC-j\omega RC}{R^{2}-\omega ^{3}R^{2}LC^{2}+j2\omega ^{2}RLC+\omega L}
\end{align*}
So, I can't separate this into its real and imaginary parts, so I can then calculate the magnitude as you described. Any ideas what I shoud do instead?

 

[The Electrician]

@The Electrician
So I tried rationalization, but ended up with an expression that has j in the denominator:
\begin{align*}
\frac{R}{R+(j\omega L)(j\omega RC+1)} \cdot \frac{R-(j\omega L)(j\omega RC+1)}{R-(j\omega L)(j\omega RC+1)} &= \frac{R^{2}-R(-\omega ^{2}RLC + j\omega L)}{R^{2}-[(-\omega L)(-\omega ^{2}R^{2}C^{2}+j2\omega RC+1)]} \\ \\
&= \frac{R^{2}+\omega ^{2}R^{2}LC-j\omega RC}{R^{2}-\omega ^{3}R^{2}LC^{2}+j2\omega ^{2}RLC+\omega L}
\end{align*}
So, I can't separate this into its real and imaginary parts, so I can then calculate the magnitude as you described. Any ideas what I shoud do instead?

You didn't do the rationalization correctly; you don't have the correct complex conjugate. You have to multiply out the two expressions in parentheses in the denominator. Multiply out (j w L)(j w R C); then you can form the correct complex conjugate.

 

[SumDood_]

@The Electrician
Tried again:
\begin{align*}

\frac{R}{R-(j\omega ^{2}RLC + j\omega L)} \cdot \frac{R-\omega ^{2}RLC-j\omega L}{R-\omega ^{2}RLC-j\omega L} &= \frac{R^{2}-\omega ^{2}R^{2}LC - j\omega RL}{(R-\omega ^{2}RLC)^{2}+(\omega L)^{2}} \\ \\
&=\frac{R^{2}-\omega ^{2}R^{2}LC}{(R-\omega ^{2}RLC)^{2}+(\omega L)^{2}}-j\frac{\omega RL}{(R-\omega ^{2}RLC)^{2}+(\omega L)^{2}}
\end{align*}

I appreciate the help, but I don't think this is how the question is supposed to be solved. I think the original approach I took was more inline with what is expected.

 

[The Electrician]

But what you were doing originally didn't work. What I'm showing you will work.

Part of what you posted in the previous post got cut off. Here's what you should get. Now set the square root of sum of squares equal to 1 and solve. It's a lot of complicated algebra. Would you be allowed to find a numerical solution?

Edit: gneill has shown how to have a lot less algebra!

 

[gneill]

Start with:
$$ \frac{V_R}{V_S} = \frac{R}{R + \left( j \omega L \right) \left(j \omega R C + 1 \right)}$$
$$\left| \frac{V_R}{V_S} \right| = \left| \frac{R}{R - \omega^2 LRC + j \omega L} \right|$$
But $\left| \frac{V_R}{V_S} \right| = 1$, so:
$$1 = \frac{R}{\left| R - \omega^2 LRC + j \omega L \right|}$$
$$1 = \frac{R}{\sqrt{ \left( R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2}}$$
$$1 = \frac{R^2}{\left( R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2} $$
Solve for C.

 

[The Electrician]

I have a small nit to pick.
In steps 3 and 4 you show the numerator as R, but for the transition from step 2 to step 3, the numerator in step 3 (and then step 4) should be |R| rather than just R. Doing this would make the step 2 to step 3 transition be true even if the numerator had been complex, with a non-zero imaginary part, and the whole procedure would be rigorously true .

Fortunately in this case it's true that |R| is identical to R, because in general (|x + jy|) is not identical to (x + jy)

 

[gneill]

I have a small nit to pick.
In steps 3 and 4 you show the numerator as R, but for the transition from step 2 to step 3, the numerator in step 3 (and then step 4) should be |R| rather than just R. Doing this would make the step 2 to step 3 transition be true even if the numerator had been complex, with a non-zero imaginary part, and the whole procedure would be rigorously true .

Fortunately in this case it's true that |R| is identical to R, because in general (|x + jy|) is not identical to (x + jy)

True. I admit I took a shortcut where $\left| R \right| = R$, since $R$ is positive real.

 

[The Electrician]

True. I admit I took a shortcut where $\left| R \right| = R$, since $R$ is positive real.

Fortunately for SumDood_ your much simpler expression will make for far less algebra in part C of his problem, what with derivatives and all.

 

[SumDood_]

You guy's help is extremely appreciated!
\begin{align*}
1 &= \frac{R^2}{\left( R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2} \\ \\
R^2 &= \left(R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2 \\ \\
R^2 &= R^{2}-2\omega^2R^2LC+\omega^4R^2L^2C^2 + \left(\omega L\right)^2 \\ \\
0 &= \omega^4R^2L^2C^2 - 2\omega^2R^2LC +(\omega L)^2
\end{align*}
After substitution:
$C = 582.911\mu F$ OR $C = 120.7077\mu F$, how do I determine which value is the correct one?

 

[The Electrician]

You guy's help is extremely appreciated!
\begin{align*}
1 &= \frac{R^2}{\left( R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2} \\ \\
R^2 &= \left(R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2 \\ \\
R^2 &= R^{2}-2\omega^2R^2LC+\omega^4R^2L^2C^2 + \left(\omega L\right)^2 \\ \\
0 &= \omega^4R^2L^2C^2 - 2\omega^2R^2LC +(\omega L)^2
\end{align*}
After substitution:
$C = 582.911\mu F$ OR $C = 120.7077\mu F$, how do I determine which value is the correct one?

They're both correct; they both satisfy the original equation.

Now can you solve part C?

 

[SumDood_]

@The Electrician
I tried this:
\begin{align*}
V_R &= V_S \cdot \frac{R}{R + \left( j \omega L \right) \left(j \omega R C + 1 \right)} \\ \\
V_R &= R\cdot V_S \cdot \left( R-\omega^2RLC+j\omega L\right)^{-1} \\ \\
\frac {dV_R} {dC} &= -R\cdot V_S \cdot \left( R-\omega^2RLC+j\omega L\right)^{-2} \left(-\omega^2RL\right)\\ \\
\frac {dV_R} {dC} &= V_S\cdot\frac{\omega^2RL}{\left( R-\omega^2RLC+j\omega L\right)^{2}} \\ \\
\text{To find the maximum:} \\
0 &= V_S\cdot\frac{\omega^2RL}{\left( R-\omega^2RLC+j\omega L\right)^{2}}
\end{align*}

Doesn't look like it is right.

 

[gneill]

I think it would be simpler to work from the magnitudes:
$$\left| \frac{V_R}{V_S} \right| = \frac{\left| R \right|}{\left| R - \omega^2 R L C + j \omega L \right| }$$
Take the magnitudes and clear the square root on the right hand side by squaring; it doesn't matter if you maximize the left side value or its square, they will give the same value for C.
$$\left| \frac{V_R}{V_S} \right| ^2 = \frac{R^2}{R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2 }$$
Find the derivative with respect to C and set it equal to zero to find the maximum. Note that you only need the numerator portion of the derivative.

 

[The Electrician]

@The Electrician
I tried this:
\begin{align*}
V_R &= V_S \cdot \frac{R}{R + \left( j \omega L \right) \left(j \omega R C + 1 \right)} \\ \\
V_R &= R\cdot V_S \cdot \left( R-\omega^2RLC+j\omega L\right)^{-1} \\ \\
\frac {dV_R} {dC} &= -R\cdot V_S \cdot \left( R-\omega^2RLC+j\omega L\right)^{-2} \left(-\omega^2RL\right)\\ \\
\frac {dV_R} {dC} &= V_S\cdot\frac{\omega^2RL}{\left( R-\omega^2RLC+j\omega L\right)^{2}} \\ \\
\text{To find the maximum:} \\
0 &= V_S\cdot\frac{\omega^2RL}{\left( R-\omega^2RLC+j\omega L\right)^{2}}
\end{align*}

Doesn't look like it is right.

The expression you got for the derivative only has "C" in the denominator. The only way that expression can be zero is if "C" is infinite. The insight you had not yet experienced way back in post #1 is that you must work with magnitudes and get rid of the imaginary parts of expressions. gneill has shown you the way.

 

[SumDood_]

$$\left| \frac{V_R}{V_S} \right| ^2 = \frac{R^2}{R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2 }$$

Let's try one more time:
\begin{align*}
\frac{d}{dC}\left| \frac{V_R}{V_S} \right| ^2 &= \frac{d}{dC}\left(\frac{R^2}{R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2 }\right) \\ \\
&=\frac{d}{dC}\left[R^2(R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2 )^{-1}\right] \\ \\
&= -R^2(-2\omega^2R^2L + 2\omega^4R^2L^2C)(R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2)^{-2} \\ \\
&= -R^2\cdot\frac{-2\omega^2R^2L + 2\omega^4R^2L^2C}{(R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2)^{2}}
\end{align*}

To find the maximum:
\begin{align*}
0 &= -R^2\cdot\frac{-2\omega^2R^2L + 2\omega^4R^2L^2C}{(R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2)^{2}} \\ \\
0 &= -2\omega^2R^2L + 2\omega^4R^2L^2C \\ \\
C &= \frac{1}{\omega^2 L} = \frac{1}{(120\pi)^2\cdot 20\text{m}} = 351.81\mu F
\end{align*}

Never have I spent this much time on one question. Thank you for helping me!
Edit: It seems I need to take advantage of working with magnitudes. Though I am not sure when is it appropriate to do so.