Is there a faster and easier way to find these values in a circuit

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In summary, the conversation discusses an easier method for finding the values of x and y in a power supply circuit. The shortcut involves converting between series and parallel configurations and using the formulas Xp = (Xs^2 + Ys^2)/Xs and Yp = (Xs^2 + Ys^2)/Ys. Other methods such as using Excel or the Smith Chart are also mentioned, but may not be allowed in exams. The conversation also touches upon the formula Xs = √ (Rp * Rs – Rs^2) used for impedance matching.
  • #1
MissP.25_5
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Is there any other easier way to get the answer for no. 2?

Power supply voltage E=8+4j [V]; Current I=1-j [A].
1) Find the power factor of the circuit as seen from the power supply.
2)Find the values of x and y.

There's a little mistake at the very last part of my working, but anyways, I want to know if there is an easier way.
 

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  • #2
I can't think of a distinctly easier method. You have two unknowns so you have to come up with two equations. Equating the terms of the known impedance to those of the expression for the impedance seems logical. After that you bring to bear any algebra tricks you've learned. (hint: consider forging a new expression by taking the ratio of the terms)
 
  • #3
gneill said:
(hint: consider forging a new expression by taking the ratio of the terms)

How to do that? :eek:
 
  • #4
MissP.25_5 said:
How to do that? :eek:

First write out the two terms (real and imaginary) equated to their know values.
 
  • #5
gneill said:
First write out the two terms (real and imaginary) equated to their know values.

That's what I did. But I still don't get the answer :/
 
  • #6
MissP.25_5 said:
That's what I did. But I still don't get the answer :/
Looking at your work, I don't clearly see the two separate equations as such. Can you write them out?
 
  • #7
gneill said:
Looking at your work, I don't clearly see the two separate equations as such. Can you write them out?

I got it! There was a lot of miscalculations before. I got x=5, y=5/2. Correct?
 
  • #8
MissP.25_5 said:
I got it! There was a lot of miscalculations before. I got x=5, y=5/2. Correct?

Your result looks fine.
 
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  • #9
gneill said:
Your result looks fine.

Thanks!I hope I pass this subject.
 
  • #10
MissP.25_5 said:
Is there any other easier way to get the answer for no. 2?

Power supply voltage E=8+4j [V]; Current I=1-j [A].
1) Find the power factor of the circuit as seen from the power supply.
2)Find the values of x and y.

There's a little mistake at the very last part of my working, but anyways, I want to know if there is an easier way.

Yes, there is an easier way.
Once you’ve divided the complex source voltage by the complex current, and subtracted the (1+j8) you have the series equivalent of the other components. This type of calculations go back and forth between series and parallel configurations, so it is a good idea to develop formulas or even functions you can use for those conversions over and over again.

Instead of using the method taught in class to multiply and divide complex numbers by converting back and forth from rectangular to polar, here is a short cut.

The series value of Xs and Ys is 1-j2.
Xp = (Xs^2 + Ys^2)/Xs = (1^2 +(- J2^2))/1 = 5
Yp = (Xs^2 + Ys^2)/Ys = (1^2 +(- J2^2))/-j2 = -j2.5

This shows the answer you arrived at for Yp should be –j2.5 instead of +j2.5.
 
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  • #11
WOW!I didn't see that earlier. That is definitely a shortcut. By the way, in the diagram it already shows the negative sign with y, so all we need is the y value, without the negative sign. But I don't understand this part:

skeptic2 said:
The series value of Xs and Ys is 1-j2.
Xp = (Xs^2 + Ys^2)/Xs = (1^2 +(- J2^2))/1 = 5
Yp = (Xs^2 + Ys^2)/Ys = (1^2 +(- J2^2))/-j2 = -j2.5

How did you get to find the parallel values like that?
 
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  • #12
I derived that about 20 years ago but I no longer remember the derivation. I haven't seen it published anywhere. Based on these formulas, can you figure out how to go from parallel to series?

Here is another method. http://www.eetimes.com/document.asp?doc_id=1278134

By the way you can use Excel as a pretty good complex arithmetic calculator, even if a little clunky. You first have to load it into Excel. To do that, click on the question mark inside the blue circle at the upper right in Excel and type "complex" into the search window and follow the instructions. This allows you to set up the formulas where all you have to do is plug in values. You can even create your own functions.

The Smith Chart or rather a version of the Smith Chart called the Immittance Chart can also do the conversion graphically. An explanation of the Smith Chart and how to use it is a little long to include here. http://www.microwavesoftware.com/manuals/imchart.html
 
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  • #13
skeptic2 said:
I derived that about 20 years ago but I no longer remember the derivation. I haven't seen it published anywhere. Based on these formulas, can you figure out how to go from parallel to series?

Here is another method. http://www.eetimes.com/document.asp?doc_id=1278134

By the way you can use Excel as a pretty good complex arithmetic calculator, even if a little clunky. You first have to load it into Excel. To do that, click on the question mark inside the blue circle at the upper right in Excel and type "complex" into the search window and follow the instructions. This allows you to set up the formulas where all you have to do is plug in values. You can even create your own functions.

The Smith Chart or rather a version of the Smith Chart called the Immittance Chart can also do the conversion graphically. An explanation of the Smith Chart and how to use it is a little long to include here. http://www.microwavesoftware.com/manuals/imchart.html

Thanks for the info, but in exams, we're not allowed to use calculators and definitely not excel. So I guess, I'll just equate the terms then. But it really got easier when we subtract the known impedance from the total impedance, rather than summing all the impedance. Thanks!
 
  • #14
I do remember how I found the formulas for series to parallel conversion. They’re nothing more than a rewriting of this formula which is used for impedance matching. I got this first formula from a Motorola (now Freescale) RF Transistor Data book. It’s amazing that as simple as this is, it’s not more widely used.

Xs = √ (Rp * Rs – Rs^2)

Rp = (Xs^2 + Rs^2) / Rs

It was probably a lucky guess that by replacing Rs with Xs would give Xp.

Impedance matching is really nothing more than series to parallel conversion. Since the components in parallel always have higher values than the equivalent impedance in series, matching is done by selecting the series reactance that will convert the lower resistance to the higher resistance in a parallel configuration. The first formula is how you calculate that series reactance.

For instance, suppose we want to match an impedance of 50 + j0 to a load of 10 + j15. First we would calculate the reactance needed to transform a series 10 + j15 to a parallel 50 +jX. We do that with the top formula and get plus or minus 20 Since we already have j15 ohms of reactance we only need j5 more. We get that by adding an inductor of j5 ohms in series.

Now we convert to parallel and get 50 + j33.33. Now all we have to do is add a shunt capacitor with a reactance of –j33.33 to cancel the +j33.33 and we have our match.

P.S. My suggestion to use a calculator or Excel to do the calculations wasn't meant for use during tests but for homework and later on the job.
 
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FAQ: Is there a faster and easier way to find these values in a circuit

How can I find the values in a circuit faster and easier?

There are a few ways to find values in a circuit faster and easier. One approach is to use a circuit simulator software, which can quickly calculate the values for you. Another method is to use a multimeter, which can measure different parameters in the circuit and help you find the values. Additionally, you can use Ohm's law and other circuit analysis techniques to calculate the values manually.

Is there a specific tool or equipment that can help me find values in a circuit faster?

Yes, there are various tools and equipment that can assist in finding values in a circuit faster. Some examples include circuit simulator software, multimeters, oscilloscopes, and signal generators. These tools are specifically designed for circuit analysis and can provide accurate and quick results.

Can I use any online resources to help me find values in a circuit?

Yes, there are many online resources available that can aid in finding values in a circuit. You can find circuit simulators, calculators, and tutorials that can guide you through the process of finding values in a circuit. However, it is essential to ensure the reliability and accuracy of these resources before using them.

Are there any shortcuts or tricks to find values in a circuit faster?

While there are no shortcuts or tricks to find values in a circuit, you can use shortcuts in circuit analysis techniques like Kirchhoff's laws and Thevenin's theorem to simplify the calculations. Additionally, using the right tools and equipment can also save time and make the process more efficient.

Can I find values in a circuit without any prior knowledge or experience?

It is possible to find values in a circuit without any prior knowledge or experience, but it is not recommended. Having a basic understanding of circuit analysis and the use of tools and equipment can help you interpret the results accurately and troubleshoot any issues that may arise. It is always best to seek assistance from a trained professional or conduct thorough research before attempting to find values in a circuit on your own.

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