Circuit help

[bobby2017]

Hi guys can someone please help?

 

[marconi]

Bobby2017; Good on you for having a bash. Quite a bit along the right lines. I am out tonight but tomorrow morning I will go through the problem with you.

 

[bobby2017]

Bobby2017; Good on you for having a bash. Quite a bit along the right lines. I am out tonight but tomorrow morning I will go through the problem with you.

Thanks very much Marconi, i appreciate it

 

[H]

What’s the answer anyway? Asking for a friend

 

[Julie.]

Ok, so since man United won I thought I would run through the calculation.

By the sounds of it, I don't think you have used complex numbers, however since it was described, I have done it both using complex numbers, and the "long-hand" way.

Option 1 just run through the method, but unless you really know how to use complex numbers I am not sure it is of any use.

Option 2, you need to cross check with your notes to see how it's done, it is straightforward but the maths looks horrible, but that's really because you have to add two vectors (actually phasers in electrical) - which is awkward to write out cleanly.

Anyway see the attached scrawl.

 

[bobby2017]

Ok, so since man United won I thought I would run through the calculation.

By the sounds of it, I don't think you have used complex numbers, however since it was described, I have done it both using complex numbers, and the "long-hand" way.

Option 1 just run through the method, but unless you really know how to use complex numbers I am not sure it is of any use.

Option 2, you need to cross check with your notes to see how it's done, it is straightforward but the maths looks horrible, but that's really because you have to add two vectors (actually phasers in electrical) - which is awkward to write out cleanly.

Anyway see the attached scrawl.

Julie you are an absolute life saver !!
I cant thank you enough!!!

 

[Lucien Nunes]

Julie has also taken the frequency as 50Hz which is not unreasonable, especially given the definition of the voltage source in the SPICE printout, but the only mention of frequency in the question itself is the statement that it is variable. Does this relate to another part of the question that we can't see?

 

[Julie.]

Julie you are an absolute life saver !!
I cant thank you enough!!!

You still have the last part to do, and take account of the power factor in the last calculation, I included the wrong photo!

The point is to understand it more than just answer the question

 

[marconi]

Adobe Document Cloud - https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A16f961cb-e305-4171-a592-78c99df6bf92

Bobby2017: Please see my working at the end of the link.

The learning points are:

1. First study and identify the circuit, so in this case the circle with a squiggle indicates it has a source of alternating current - 240Vrms and 'variable frequency' - we will assume 50Hz for our solution.

2. There are two branches in parallel. Indicate on the diagram the currents as I have with I, i1 and i2. We also note that since these branches are in parallel the same emf is across both viz the 240V ac supply. The current through each is only dependent on the voltage and frequency of the generator and not the current in the other branch.

3. Mark up the components with their values in Ohms, Henry and Farad and label the components as C, L and R. The next step (top right) is to represent these components by their resistance and reactance XC, XL and R. You can see my working for Xc and XL. Reactance has the same units as resistance.

4. Then we redraw the circuit with the resistances and reactances written beside each components but the components are now generalised impedances (the rectangular boxes) Because we have stopped using the symbols for a capacitor and inductor we need to indicate the nature of the impedance by including the j operator; plus j for an inductance and minus j for a capacitor.

5. For the series combination of XL and R we combine their impedances just as we would combine pure resistances to arrive at ZLR.

6. I have redrawn the circuit to show ZC and ZLR.

7. Now we apply Ohm's Law but generalised for ac circuits ... V = I x Z, so I = V/Z etcetera. This enables us to work out i1 and i2 and thence I.

8. We do the complex number algebra and end up with the currents in the form a +jb. In electrical science this means that the current has a phase and a quadrature components which in plainer language means a componet in phase with the applied voltage and a components at 90 degrees in phase to the voltage - the sign +/- indicates if the quad component leads or lags. One can convert Cartesian to polar(Modulus and Argument) using standard mathematics - look this up by googling 'Complex Number Algebra'.

9. 'Looking out' from the generator towards the two branches, the generator 'sees' a load with a voltage V across it and drawinf a current I. The load's effective or equivalent impedance is V/I - Zequiv. AS this is in Cartesian form one can readily decompose it into its constituent components as pure resistance and pure reactance as I show with R' and L'. It is an inductance because j is positive; had the reactance been negative I would have shown a capacitor. The effective impedance then is inductive in nature but with some resistance too. One can fund out the value of the effective inductance or capacitance using the formulas for XC and XL.

10. The last question is about how much power is consumed by the circuit. This question is rather loosely defined because in ac circuits we speak of apparent, real(or active) and reactive power. Real or active power is what does work or Joule/Ohmic heating whereas reactive power is that which flows back and forth between the generator and the reactances of the circuit. Apparent power is simply the voltage across the load/circuit times the current flowing through it. You can look up for yourself some further explanations of these terms. Joule/Ohmic heating in this circuit only takes place in the resistor and the power dissipated in a resistor is easily calculated since you know the value of the R' and the total current I. You could also calculate the power using two different values for current and resistance - which two are they?

11. For the evaluation question one would perhaps mention additionally the trigonometric approach which uses so-called phasors whose lengths and directions represent the complex quantities a+bj; you can look up this method here:

https://www.engr.siu.edu/staff2/spezia/Web332b/Lecture Notes/Lesson 2_et332b.pdf

One could measure the current I as the frequency of the generator was varied from 0 to 50Hz and plot this current against frequency to show the 'frequency response' of the circuit. By measuring the phase difference between the voltage sinusoid and the current sinusoid for each spot frequency one can also plot phase shift versus frequency. These two plots - called a Bode plot fully represent the electrical characteristics of the total circuit.

http://www.dartmouth.edu/~sullivan/22files/Bode_plots.pdf

And then there is the sue of computer modelling of electrical circuits using SPICE -

SPICE - Wikipedia - https://en.wikipedia.org/wiki/SPICE

I think this last part is beyond the scope of your current state of knowledge.

Anyway, enjoy your new found proficiency with ac circuit analysis.


M

 

[marconi]

Bobby2017: I missed out a fourth power sometimes but not often found in electrical engineering - girlpower (see Julie #34) - strangely its power factor is greater than 1.

 

[telectrix]

once was able to do these sort of calcs, but last time was 1970, these days i've got more chance of flying to the moon.

 

[marconi]

once was able to do these sort of calcs, but last time was 1970, these days i've got more chance of flying to the moon.

Now you gift the wisdom of your experience.