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I am trying to get some practice on the basics, but i am having some issues with the following question:

Inductor, Resistor Series/Capacitor Parallel

  1. Find Currents I(1) and I(2) and show them on a phasor diagram.
  2. Find the total current, I(total).
  3. Find the Voltage across each component.
  4. Calculate the total circuit impedance, Z(total).

Capacitor, Resistor Series/Inductor Parallel

  1. Find the total current I(total).
  2. Find the voltage across each component.
  3. Calculate the total circuit impedance Z(total).

For #1, I have worked out the answer for I(1) to be 3.88 A and I(2) to be 2.26 A. Does this seem correct?

For #2, I have worked out an answer of 2.59 A for I(total). But, this cannot be correct, right?

I have looked online for guidance and it seems it may be useful to use complex numbers for working it out, but I do not know where this would fit in, or even if this is required to answer it.

Any help or point in the right direction would be appreciated. Thanks

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  • \$\begingroup\$ Okay you said you're a little confused on complex numbers. Yes, I would use complex numbers to solve for the current because it's just easier math. You know that the impedence of a capacitor is \$ \displaystyle Z_C =-j\omega C= \frac{1}{j\omega C}\$ and for an inductor it's \$Z_L= j\omega L\$. You should treat them as if they were resistors but with complex numbers instead then apply Ohm's Law like normal. You will have to do deal Kirchhoff's laws as well. \$\endgroup\$ – KingDuken Aug 2 '18 at 21:12
  • \$\begingroup\$ Your two currents in #1 seem right, but you haven't included the angles. Since you write "show them on a phasor diagram" it seems to me you need to add that. You'll need that information to work out the total current for #1. For #2, you did not get the right total current. Perhaps because you aren't taking angles into account? \$\endgroup\$ – jonk Aug 2 '18 at 21:14
  • \$\begingroup\$ Take #1. You can easily work out the value for \$X_C\$ given the frequency. From that, you can work out the current magnitude in that branch. The current angle leads and will be +90 degrees. For the other leg of #1, with an inductor and resistor, you can work out the impedance there knowing that the resistor's phase angle is 0 degrees and that the inductor's reactance, \$X_L\$, will be -90 degrees. If you know your side-angle-side triangle formulas, you can work out all the details without complex numbers. But complex numbers help a lot, once you get used to them. \$\endgroup\$ – jonk Aug 2 '18 at 21:34
  • \$\begingroup\$ Hi @jonk, thanks for taking the time to help me with this! I have approached this without using complex numbers (as I am not well-versed in using them - I have had very little practice with them in the past); i have answered questions 1 - 3. With #4, what is the impedance equation needed/how do i derive this? Due to the circuit being a mix of Series and Parallel components is there an easy way of adding the component impedances to attain the Total Impedance? Apologies if this seems a bit basic! I am just starting out and am trying to muddle my way through \$\endgroup\$ – Tay1995 Aug 3 '18 at 18:26
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    \$\begingroup\$ @Tay1995 If you decide you want to "do some math learning" as an aside to electronics (you know, take a small "detour"), I'd highly recommend going to youtube and looking up a content provider who goes by the name 3blue1brown. I can't think of a better set of videos to study. He includes Fourier, zeta, complex numbers, and a host of other things. But he focuses on making these things profoundly visual and instinctual. Worth every moment you can afford. (Way better than Khan academy, for what he does do.) \$\endgroup\$ – jonk Aug 3 '18 at 19:58
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This is pretty basic stuff and you have to understand them well in order to proceed in mode practical (and complicated) circuits. Solving it for you, would removed the meaning of the exercise, but here are some basic tools:

Ohm's Law

\$V = I \cdot Z\$, where Z is the impedance (\$Z = \sqrt{ (R^2 + X^2}\$)

for inductors $$X_L = j \cdot ω \cdot L,\;\; ω = 2\cdot \pi \cdot f$$

and for capacitors $$X_C = \frac{1}{j \cdot ω \cdot C} = -\frac{j}{ω \cdot C}$$

For example, R and L are in series so:

$$Z_1 = R + jwL = 40 + j47.1 = 61.8\angle49.66$$

$$I1 = \frac{V}{Z_1} = \frac{240\angle0} {61.8\angle{49.66}} = 3.88\angle-49.66$$

Doing the rest is trivial.


EDIT: just to clarify, phasors can have two forms

  1. Cartersian: \$Z = R + jX\$
  2. Polar: \$Z = |Z|\angle φ\$

Transformations from one form to the other are known from vector analysis

$$ |Z| = \sqrt{R^2 + X^2} $$ $$ φ = atan2(\frac{X}{R}) $$

and

$$R = |Z|\cdot cos(φ)$$ $$X = |Z|\cdot sin(φ)$$

There is a plethora of videos on youtube to learn more about phasors and vector analysis, which also include animations

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No complex numbers below.

Reactive Components

When you think of a capacitor, just remember that the voltage across the capacitor takes time to increase or decrease as current into or out of the capacitor is applied. This means that for capacitors, you can either say, "voltage lags the current by \$90^\circ\$" or else "current leads voltage by \$90^\circ\$." Either statement is correct. It's just two different ways of saying the same thing.

When you think of an inductor, just remember that the current through the inductor takes time to increase or decrease as voltage across it is applied (the direction of the change in current depending on the polarity of the applied voltage.) This means that for inductors, you can either say, "current lags the voltage by \$90^\circ\$" or else "voltage leads current by \$90^\circ\$." Either statement is correct. It's just two different ways of saying the same thing.

Your first circuit includes two parallel branches. One has only a capacitor. The other has a resistor and an inductor, in series. It's easiest if you treat these branches separately and then combine the results to get the total current.

For your first circuit, the two reactances are:

$$\begin{align*} X_\text{C}=\frac{1}{2\pi f \:C}&\approx 106.1\:\Omega\\\\ X_\text{L}=2\pi f\: L&\approx 47.12\:\Omega \end{align*}$$

Series Resistor and Inductor

Here is a series branch that includes a resistor and an inductor:

enter image description here

You can see here that the direction of both the current and voltage for the resistor are the same. Note that isn't true for the inductor. Since this is a series branch, the current in both components must be the same and in the same direction. So I've shown this fact by having both current arrows pointing in the same direction. However, the inductor's voltage leads the current by \$90^\circ\$. So it's voltage arrow is shown with a positive angle (based upon conventional trigonometry and the unit circle.)

The current is the same in both, but the voltage across each is oriented at different angles. So let's focus on the resulting voltage diagram:

enter image description here

The left side follows the earlier diagram's orientation for voltages and shows you what the resulting applied voltage vector looks like. You know the applied voltage vector's magnitude, already. It's \$240\:\text{V}\$. It's the other two voltages that you don't know, just yet. But at least you can see their relationship.

The right side rotates the left diagram so that the voltage now occupies the \$x\$-axis, so that the voltage is taken to be, arbitrarily, at \$0^\circ\$. As you can see, this means that the current through the series combination will have a negative angle relative to the voltage (the "reference.") Also take note that the current arrow is parallel to the resistor's voltage arrow. So you know that the resistor's voltage will have the exact same angle as the current.

(Of course, if you are supplied with a problem that specifies an angle for the applied voltage, then you might rotate this so that the applied voltage isn't on the \$x\$-axis as shown here but instead at whatever angle is given to you.)

Now, looking at the above diagrams, imagine what happens as either \$R\$ or \$L\$ disappears. If \$L\$ were absent, then \$V\$, \$V_\text{R}\$, and \$I\$ would all line up together. No angles to worry about. If \$R\$ were absent, then \$V\$ and \$V_\text{L}\$ would line up together, but would be perpendicular to \$I\$ and leading the current by \$90^\circ\$. (Just imagine varying each and see how the diagrams rotate and change as you increase or decrease the impedance of either series component.)

Series Resistor and Capacitor

Here is a series branch that includes a resistor and a capacitor:

enter image description here

Again, you can see that the direction of both the current and voltage for the resistor are the same. And again, you note that isn't true for the capacitor. Since this is a series branch, the current in both components must be the same and in the same direction. Again, I've shown this fact by having both current arrows pointing in the same direction. However, the capacitor's voltage lags the current by \$90^\circ\$. So it's voltage arrow is shown with a negative angle (based upon conventional trigonometry and the unit circle.)

The current is the same in both, but the voltage across each is oriented at different angles. So let's focus on the resulting voltage diagram:

enter image description here

Again, the left side follows the earlier diagram's orientation for voltages and shows you what the resulting applied voltage vector looks like. Again, you know the applied voltage vector's magnitude, already. It's \$240\:\text{V}\$. And again, it's the other two voltages that you don't know, just yet. But at least you can see their relationship.

The right side rotates the left diagram so that the voltage now occupies the \$x\$-axis, so that the voltage is taken to be, arbitrarily, at \$0^\circ\$. As you can see, this means that the current through the series combination will have a positive angle relative to the voltage (the "reference.") Also again take note that the current arrow is parallel to the resistor's voltage arrow. So you know that the resistor's voltage will have the exact same angle as the current.

Now, looking at the above diagrams, imagine what happens as either \$R\$ or \$C\$ disappears. If \$C\$ were absent, then \$V\$, \$V_\text{R}\$, and \$I\$ would all line up together. No angles to worry about. If \$R\$ were absent, then \$V\$ and \$V_\text{C}\$ would line up together, but would be perpendicular to \$I\$ and lagging the current by \$90^\circ\$. (Just imagine varying each and see how the diagrams rotate and change as you increase or decrease the impedance of either series component.)

Current and Impedance of the Left Leg of the First Circuit

Let's start with the left side branch, composed of \$R\$ and \$X_\text{L}\$, using what's discussed above. I'm going to snap a copy of part of the image and add some text:

enter image description here

The three purple arrows form up a right-triangle, with \$I\cdot Z\$ representing the hypotenuse. But you already know the magnitude here; it's \$240\:\text{V}\$. How do you figure out the other two magnitudes? It's not so hard. Just use the basic Pythagorean theorem:

$$\begin{align*} \left(I\cdot Z\right)^2&=\left(I\cdot X_\text{L}\right)^2+\left(I\cdot R\right)^2\\\\ \left(240\:\text{V}\right)^2 &=I^2\cdot\left[\left(47.12\:\Omega\right)^2+\left(40\:\Omega\right)^2\right]\\\\\therefore\\\\ I &= \frac{240\:\text{V}}{\sqrt{\left(47.12\:\Omega\right)^2+\left(40\:\Omega\right)^2}}\approx 3.883\:\text{A} \end{align*}$$

From that, you can compute \$Z=\frac{240\:\text{V}}{3.883\:\text{A}}\approx 61.81\:\Omega\$. (Just looking at the above equation, I think you can also see that you can also compute it as \$Z=\sqrt{R^2+X_\text{L}^2}\$.)

The angle for the current, with respect to the applied voltage, is negative and can be computed as \$-\operatorname{atan}\left(\frac{X_\text{L}}{R}\right)\$; or about \$-49.67^\circ\$.

(Note that the above is only for the left series leg in your first circuit. The capacitor hasn't yet been added to the picture.)

Now, while you know the angle for the current with respect to the voltage (treated as \$0^\circ\$), the angle for the impedance \$Z\$ is taken with respect to the current. So in this case the impedance angle is positive, not negative. But it is the same magnitude. (Just keep in mind what question is being asked when you decide which sign to use.)

You can also now compute the individual magnitudes for the voltage across the resistor and the voltage across the inductor by multiplying each individual impedance by the computed branch current. And you can easily work out the angles, too, by simply looking at the diagram and using the angle computed for the current with respect to the voltage.

Current and Impedance of the Right Leg of the First Circuit

The capacitor leg is easier and I won't bother with a picture, this time. The current magnitude is \$\frac{240\:\text{V}}{106.1\:\Omega}\approx 2.262\:\text{A}\$. And you should now already know the angle for the current, with respect to the applied voltage for a capacitor. (If not, review near the top above.)

I'm now going to jump forward to the total impedance question.

Total Circuit Impedance

The remaining problem is in figuring out the total impedance of your first circuit. To achieve that, you need to combine the currents of each of the two legs, properly.

Before I dive in, let me remind you of something. The exact same voltage is applied to both legs. So our problem isn't figuring out the applied voltage. That's already known in advance. Our problem is in combining the two currents. So follow along closely. Once we have the total current, the total impedance is pretty easy.

The following diagram shows the total voltage that is in common to both legs as being the same arrow at \$0^\circ\$; both on the left and on the right:

enter image description here

On the left side, you can see the capacitor's current oriented correctly with respect to the applied voltage (in a capacitor, the current leads the voltage.) You can also see the resulting current for the resistor in series with the inductor, too, also oriented with the angle we computed earlier.

On the right side, I've simply moved the capacitor's current to the tip of the series inductor+resistor current, so that they are summed. You can see the result of this sum as the green arrow on the right side diagram.

To sum up these currents, you might start by using a SAS (side-angle-side) triangle computation: \$c=\sqrt{a^2+b^2-2\:a\cdot b\cdot\operatorname{cos}\left(\theta\right)}\$. You know the magnitude of \$I_{\text{R}+\text{L}}\$ and \$I_\text{C}\$ and you can work out the angle between them easily enough, as well. So you can compute \$c\$ for \$I_\text{TOT}\$.

And with all three sides and an angle already in hand, you can work out the remaining angles. I will let you work on the details here. But the results should be something like \$2.608\:\text{A}\quad\angle -15.53^\circ\$.

With that information, you can work out the magnitude of \$Z_\text{TOT}=\frac{240\:\text{V}}{2.608\:\text{A}}\approx 92.03\:\Omega\$. And, of course, the angle of the impedance (taken with respect to the current, as usual) is positive, not negative, and therefore expressed fully in polar form is \$Z=92.03\:\Omega\quad\angle +15.53^\circ\$.

Now take a crack at #2 and see if you can get where you need to be, there.

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  • \$\begingroup\$ This can work, but it's really way easier to understand that any vector can be written in a form \$ \vec{v} = a + jb = |v| \angle{φ} \$, since it will get more complicated as more components appear. Also, I think that the initial question is a school or college exercise and I highly doubt that this is what the expect for an answer :P \$\endgroup\$ – thece Aug 4 '18 at 9:18
  • \$\begingroup\$ @thece I can't say what the OP wanted. Only that they aren't familiar with the use of complex numbers. Also, different people have different ways of learning. Some prefer algebra, but some require a more visual approach. Who knows for sure which is best in this case? I decided to try a different approach. \$\endgroup\$ – jonk Aug 4 '18 at 14:34
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Well, first of all we can write:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{\frac{1}{\text{R}+\text{j}\omega\text{L}}+\frac{1}{\left(\frac{1}{\text{j}\omega\text{C}}\right)}} =\frac{\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}+\left\{\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}+\omega\text{C}\right\}\cdot\text{j}}{\left(\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)^2+\left(\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}+\omega\text{C}\right)^2}\tag1$$

So, when \$\omega=2\pi\cdot50=100\pi\space\text{rad/s}\$, \$\text{R}=40\space\Omega\$, \$\text{C}=30\cdot10^{-6}\space\text{F}\$ and \$\text{L}=150\cdot10^{-3}\space\text{H}\$ we can write:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1600000}{40000-3024\pi^2+81\pi^4}-\frac{3000\pi\left(9\pi^2-136\right)}{40000-3024\pi^2+81\pi^4}\cdot\text{j}\tag2$$

Finding \$\text{I}_{\space\text{total}}\$, we have to find:

$$\text{I}_{\space\text{total}}=\frac{240}{\left|\underline{\text{Z}}_{\space\text{in}}\right|^2}=\frac{6}{25}\cdot\sqrt{\frac{65600}{64+9\pi^2}+9\pi^2-400}\approx2.60786\space\text{A}\tag3$$

For \$\text{I}_1\$ and \$\text{I}_2\$ we get:

  • $$\text{I}_1=\left|\frac{\frac{1}{\text{j}\omega\text{C}}}{\text{R}+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\right|\cdot\text{I}_{\space\text{total}}=\frac{48}{\sqrt{9\pi^2+64}}\approx3.88277\space\text{A}\tag4$$
  • $$\text{I}_2=\left|\frac{\text{R}+\text{j}\omega\text{L}}{\text{R}+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\right|\cdot\text{I}_{\space\text{total}}=\frac{18\pi}{25}\approx2.26195\space\text{A}\tag5$$
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