Post your calculations, and the results you get, and we'll sober up a bit and see if you are right.
Hi i need help with what equations to use to get the results. I literally have 0 knowledge about this kind of stuff
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Hi i dont have any calculations because i dont know what equations to use or how to go about getting the results
If you’ve not been taught it it’s unreasonable and unrealistic if the course provider to expect you to understand it mate, there are plenty of people on here that could give you the answer....but would that really help you? You need to speak to the tutor mate...there’s no shame in putting ya hand up and saying you don’t get it, after all I assume you or someone else is paying these people to teach you it, when I did my courses it cost thousands and there was no way I was getting fobbed it till they’d taught me it ?
to be fair my tutor as about as helpful as a chocolate fireguard but thats another story. is there an equation or explanation for this that i might be able to follow? you dont have to tell me the answer, just how to do it and then i could try and do the maths on my own.
thanks in advance.
Let's start you off then.
Get your pencil.
Name the voltage where the resistors join (for the sake of argument call it V2).
Name the current in the left hand resistor I1 and in the right hand resistor I2.
Remember Ohm's law. That's it, it really is. - But to take you one step further..
Mr Georg Ohm says that I1 = (12-V2)/1000
and also that I2 = (5-V2)/1500
He also says that (I1+I2) =V2/680
The rest is up to you. Some very basic manipulation, solve for V2, I1 and I2.
Get your pencil.
Name the voltage where the resistors join (for the sake of argument call it V2).
Name the current in the left hand resistor I1 and in the right hand resistor I2.
Remember Ohm's law. That's it, it really is. - But to take you one step further..
Mr Georg Ohm says that I1 = (12-V2)/1000
and also that I2 = (5-V2)/1500
He also says that (I1+I2) =V2/680
The rest is up to you. Some very basic manipulation, solve for V2, I1 and I2.
The question asks for the solution to be found using Kirchoff's laws (presumably to demonstrate an understanding of them.
Kirchoof's current law states that the sum of currents flowing into a circuit node (a point where things connect together) must equal the sum of currents flowing out (i.e. the algebraic sum must be zero).
Kirchoff's voltage law states that the algebraic sum of voltages around a closed loop must be zero.
Take some time to internalise these concepts: they are important to understanding how things behave. After a while they will seem intuitive and you will wonder why Kirchoff got any recognition for stating the totally obvious.
Look at the loop V1-R1-R3. We know V1, and from KVL we know that the sum of the voltages across R1 and R3 must equal V1 in magnitude (taking care with the sign in each case!) The same is true of the other loops: V2-R3-R2 and V1-R1-R2-V2.
From KCL you can say things about the current in each resistor in terms of the currents in the others. As we know the resistance values, we can also write the voltages in terms of the currents through each resistor using Ohm's law. Hence we can write a bunch of simultaneous equations to solve for any of the unknowns.
Just watch the polarities / signs: Decide on a direction and stick to it even if you realise the current might be flowing the other way after all. It doesn't matter provided you are consistent.
Kirchoof's current law states that the sum of currents flowing into a circuit node (a point where things connect together) must equal the sum of currents flowing out (i.e. the algebraic sum must be zero).
Kirchoff's voltage law states that the algebraic sum of voltages around a closed loop must be zero.
Take some time to internalise these concepts: they are important to understanding how things behave. After a while they will seem intuitive and you will wonder why Kirchoff got any recognition for stating the totally obvious.
Look at the loop V1-R1-R3. We know V1, and from KVL we know that the sum of the voltages across R1 and R3 must equal V1 in magnitude (taking care with the sign in each case!) The same is true of the other loops: V2-R3-R2 and V1-R1-R2-V2.
From KCL you can say things about the current in each resistor in terms of the currents in the others. As we know the resistance values, we can also write the voltages in terms of the currents through each resistor using Ohm's law. Hence we can write a bunch of simultaneous equations to solve for any of the unknowns.
Just watch the polarities / signs: Decide on a direction and stick to it even if you realise the current might be flowing the other way after all. It doesn't matter provided you are consistent.
Kirchhoff is really "common sense" applied to Ohm's law. Once you get that it's all the same thing. Tutors may get a bit ---- about it so I understand your point. In this case Kirchhoffs "laws" are used in the way I did it also. (Not being a law at all really- just being a rather obvious observation resulting from Ohm) :
Current into a node = currents out. Hence currents I1 and I2 are summed in the middle resistor which takes I1+I2 (Kirchhoffs first law).
Voltage round a loop equals zero. This is implicit in the first and second and third statements. Insofar as I said:
I1=(12-V2)/1000 . The voltage across the battery is 12V which equals the 12 Volts across the remainder of the loop of 1000 and 680 ohm resistors. Ie 12-12 = 0 (Kirchhoff second law)
Likewise the second. The voltage across the battery is 5V which equals the 5 volts across the 1500 and 680 ohm resistors. Is 5-5 = 0 (Kirchhoff)
Another way of doing the same as Lucien points out might be to set it out more explicitly as your tutor may define as “Kirchhoff” by setting out two clockwise circulating currents (one will be negative), but at the end of the day it’s all Ohms law!
Have fun
Bobby2017: Well done for having a go. You know how to solve a problem using simultaneous equations. Alas, the simultaneous equations you wrote down are incorrect. You made errors because you did not define the currents i1 and i2 and their directions nor the voltage drops across the resistors and their polarity. This meant that when you added up the voltage drops in a loop to equal the electromotive force in the loop you had some signs wrong in the simultaneous equations.
An important lesson then is to define your currents and voltage drops on a diagram as I did top left in the attachment.
Normally in a dc circuit one would use conventional current flow ie from positive to negative. So current 'leaves' the positive terminal of the battery. The polarity of the voltages across a resistor when a current flows through them as I have defined them is as shown by the arrows in my attached diagram where the arrow tip is +ve with respect to its tail - note well that for the currents I have defined the voltage drop polarity (arrow direction) is in the opposite direction.
Applying Kirchoff's voltage law = Sum of emfs in a loop equals sum of voltage drops - for the left hand loop - it would have been helpful if I had remembered to put arrows beside each battery to indicate the direction of the emfs of the batteries. For the 12V battery it would point upwards and for the 5V battery it would point downwards:
12 = 1000i1 + 680(i1-i2)
For the right hand loop:
5 = 1500i2 - 680(i1-i2).......Note well the minus sign before the 680 Ohm resistor because the polarity of the voltage drop I defined across this resistor is in the opposite direction to the voltage drop across the 1500 Ohm resistor.
After deriving the simultaneous equations I then use the matrix algebra method to do the mathematics to arrive at i1 = 9.2375mA and i2 = 5.175mA. Because these are positive values these currents are indeed in the directions I defined in the diagram. A negative sign would reverse the direction of current flow.
One then finds the voltage drop across the 680 Ohm using using Kirchoff's current law. The current through the 680 Ohm resistor is i1-i2, so one can then use Ohm's Law to calculate the voltage drop. And since i1 is greater than i2, (i1-i2) is a positive number which means that the voltage drop polarity across the 680 is indeed in the direction I initially defined. If (i1-i2) was negative this would indicate the voltage drop polarity is opposite to what I defined.
As clear as mud? I hope not.
An important lesson then is to define your currents and voltage drops on a diagram as I did top left in the attachment.
Normally in a dc circuit one would use conventional current flow ie from positive to negative. So current 'leaves' the positive terminal of the battery. The polarity of the voltages across a resistor when a current flows through them as I have defined them is as shown by the arrows in my attached diagram where the arrow tip is +ve with respect to its tail - note well that for the currents I have defined the voltage drop polarity (arrow direction) is in the opposite direction.
Applying Kirchoff's voltage law = Sum of emfs in a loop equals sum of voltage drops - for the left hand loop - it would have been helpful if I had remembered to put arrows beside each battery to indicate the direction of the emfs of the batteries. For the 12V battery it would point upwards and for the 5V battery it would point downwards:
12 = 1000i1 + 680(i1-i2)
For the right hand loop:
5 = 1500i2 - 680(i1-i2).......Note well the minus sign before the 680 Ohm resistor because the polarity of the voltage drop I defined across this resistor is in the opposite direction to the voltage drop across the 1500 Ohm resistor.
After deriving the simultaneous equations I then use the matrix algebra method to do the mathematics to arrive at i1 = 9.2375mA and i2 = 5.175mA. Because these are positive values these currents are indeed in the directions I defined in the diagram. A negative sign would reverse the direction of current flow.
One then finds the voltage drop across the 680 Ohm using using Kirchoff's current law. The current through the 680 Ohm resistor is i1-i2, so one can then use Ohm's Law to calculate the voltage drop. And since i1 is greater than i2, (i1-i2) is a positive number which means that the voltage drop polarity across the 680 is indeed in the direction I initially defined. If (i1-i2) was negative this would indicate the voltage drop polarity is opposite to what I defined.
As clear as mud? I hope not.
Attachments
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Thankyou so much for your help ? you are a legend !!Bobby2017: Well done for having a go. You know how to solve a problem using simultaneous equations. Alas, the simultaneous equations you wrote down are incorrect. You made errors because you did not define the currents i1 and i2 and their directions nor the voltage drops across the resistors and their polarity. This meant that when you added up the voltage drops in a loop to equal the electromotive force in the loop you had some signs wrong in the simultaneous equations.
An important lesson then is to define your currents and voltage drops on a diagram as I did top left in the attachment.
Normally in a dc circuit one would use conventional current flow ie from positive to negative. So current 'leaves' the positive terminal of the battery. The polarity of the voltages across a resistor when a current flows through them as I have defined them is as shown by the arrows in my attached diagram where the arrow tip is +ve with respect to its tail - note well that for the currents I have defined the voltage drop polarity (arrow direction) is in the opposite direction.
Applying Kirchoff's voltage law = Sum of emfs in a loop equals sum of voltage drops - for the left hand loop - it would have been helpful if I had remembered to put arrows beside each battery to indicate the direction of the emfs of the batteries. For the 12V battery it would point upwards and for the 5V battery it would point downwards:
12 = 1000i1 + 680(i1-i2)
For the right hand loop:
5 = 1500i2 - 680(i1-i2).......Note well the minus sign before the 680 Ohm resistor because the polarity of the voltage drop I defined across this resistor is in the opposite direction to the voltage drop across the 1500 Ohm resistor.
After deriving the simultaneous equations I then use the matrix algebra method to do the mathematics to arrive at i1 = 9.2375mA and i2 = 5.175mA. Because these are positive values these currents are indeed in the directions I defined in the diagram. A negative sign would reverse the direction of current flow.
One then finds the voltage drop across the 680 Ohm using using Kirchoff's current law. The current through the 680 Ohm resistor is i1-i2, so one can then use Ohm's Law to calculate the voltage drop. And since i1 is greater than i2, (i1-i2) is a positive number which means that the voltage drop polarity across the 680 is indeed in the direction I initially defined. If (i1-i2) was negative this would indicate the voltage drop polarity is opposite to what I defined.
As clear as mud? I hope not.
Bobby2017: Indeed there are many generous folk in the EF but they tend to baulk at doing other's homework. This is the second electrical science problem you have had no inkling about how to proceed. Does your tutor know of your difficulty? You said you are reading mechanical engineering; nevertheless these sorts of problems you ought to be able to tackle to be a rounded engineer.
Please let us know why you (or the whole class) have been set problems you cannot answer.
Please let us know why you (or the whole class) have been set problems you cannot answer.
What courses are you doing exactly @bobby2017?.....and what is your background that got you to this point?.....I’m not judging as I couldn’t answer the questions myself having never tried/looked before, but I find it a little bemusing that someone has found themselves in the situation you’re in and not have even an inkling of where to start, as before not judging or being a critic that’s not me I admire anyone that gets up and tries to better emselves but this thread is making me curious.....
I would have thought it's a first degree course, electrical students do a lot of mechanical stuff in the first year and conversely mechanical engineering students do a similar amount of electrical stuff.
These are the sorts of questions from basic circuit theory
To the op:
You usually get recommended a book or two per subject, I would suggest you do invest in one for this subject if you are struggling, from my experience you don't get much help from the lecturers as they expect you to be able to learn it
Back in the '70s the on-campus bookshop would sell 2nd hand books from the previous year's students, or the su would run a similar scheme - once you move on you sell the books back.
Don't know if that's the case now.
If the tutor is expecting you to know these answers, then maybe HE is in the wrong classroom.
Hi its not just me in my class that is struggling, its pretty much the whole class that is struggling, we have had a "stand in" teacher who isn't really the best of help to be honest. we have raised this as a class to our course leader, but his response was that they are struggling to recruit a good permanent electronics teacher. So for the mean time, we can either make do with what we have or go without a teacher (both are just as bad as each other by the way). So its a lose lose situation. On the brighter side for us, with no disrespect meant for anyone, is that we only have this 1 assignment for electronics and this is the final question on the assignment. I think i have managed the rest of the assignment with a lot of research but these circuits just go right over my head.
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Hi its not just me in my class that is struggling, its pretty much the whole class that is struggling, we have had a "stand in" teacher who isn't really the best of help to be honest. we have raised this as a class to our course leader, but his response was that they are struggling to recruit a good permanent electronics teacher. So for the mean time, we can either make do with what we have or go without a teacher (both are just as bad as each other by the way). So its a lose lose situation. On the brighter side for us, with no disrespect meant for anyone, is that we only have this 1 assignment for electronics and this is the final question on the assignment. I think i have managed the rest of the assignment with a lot of research but these circuits just go right over my head.
Do you know how to work out the total resistance of two resistors in series or in parallel? If you do then you can do the same for impedances - see a little further down this post.
Also do you know:
1. that the current(I) through an inductor lags the voltage (V) across it? So one can say the voltage V = jX x I where j is square root of -1 or a phase shift of 90 degrees and X is the reactance of the inductor at the frequency of the voltage V?
2. For a capacitor, V = -jX x I where this time -j means a phase shift of -90 degrees - the current leads the voltage this time.
3. that the current through a resistor is in-phase with the voltage across it so V = R x I? So no j required.
4. the reactance X of an inductor is 2 x pi x f x L where f is frequency and L is the inductance of the inductor.
5. For a capacitor X = 1/(2pi x f x C) where C is the capacitance of the capacitor.
6. That in alternating current circuits we deal in impedance which is an amalgam of resistance and the two types of reactance viz inductive and capacitive.
7. So a capacitor C in series with a resistor R would have an impedance
Z = R - j/(2pi x f x C).
8. An inductor L in series with a resistor R and capacitor C would have an impedance of
Z = R + j(2pi x F x L) - j/(2pi x f x C)
9. In general, Z = R + j(Xind - Xcap) and if the applied voltage V across an impedance Z is V = Z x I. Z is a complex number. I is the current through the impedance.
10. If one works out the current I by dividing V/Z = a +bj then the amplitude of the current is the modulus of a + bj and the phase of the current with respect to the voltage is the argument of a + bj.
Have a go using these cues to tackle the problem you have been set.
Also do you know:
1. that the current(I) through an inductor lags the voltage (V) across it? So one can say the voltage V = jX x I where j is square root of -1 or a phase shift of 90 degrees and X is the reactance of the inductor at the frequency of the voltage V?
2. For a capacitor, V = -jX x I where this time -j means a phase shift of -90 degrees - the current leads the voltage this time.
3. that the current through a resistor is in-phase with the voltage across it so V = R x I? So no j required.
4. the reactance X of an inductor is 2 x pi x f x L where f is frequency and L is the inductance of the inductor.
5. For a capacitor X = 1/(2pi x f x C) where C is the capacitance of the capacitor.
6. That in alternating current circuits we deal in impedance which is an amalgam of resistance and the two types of reactance viz inductive and capacitive.
7. So a capacitor C in series with a resistor R would have an impedance
Z = R - j/(2pi x f x C).
8. An inductor L in series with a resistor R and capacitor C would have an impedance of
Z = R + j(2pi x F x L) - j/(2pi x f x C)
9. In general, Z = R + j(Xind - Xcap) and if the applied voltage V across an impedance Z is V = Z x I. Z is a complex number. I is the current through the impedance.
10. If one works out the current I by dividing V/Z = a +bj then the amplitude of the current is the modulus of a + bj and the phase of the current with respect to the voltage is the argument of a + bj.
Have a go using these cues to tackle the problem you have been set.
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The real clue is in the question, it asks for a series of answers, if you answer each one in order, they lead on to the next.
So I just asks for the current in one branch, so from ohms law I = V /Z, you know V, and you know Z as it's R+jX, X being 2 pi f L
This will be a complex number, so you can convert it to polar to get magnitude and phase angle.
Part II asks for the current in the other branch, so repeat the above remembering that capacitors lead and the calculation for Z is slightly different to an inductor.
Part III asks for the total current, so add the two branches, either via cartesian form as you have calculated above, or even in their polar form noting the sign /direction of the complex part.
Part IV is the impedance, so you could calculate it via standard series/parallel method (using the complex values), or as this has lead you down a simpler path Z=V/I - since you have V and the calculation of total current from part III - this would be an easy way to get the complex impedance
And so on...
I don't know how far you are down your course, but in any ac system calculations the use of complex numbers is pretty standard in anything but the simplest problem.
What level of course are you doing?
This is important as the more basic courses won't use complex numbers, and the more advanced would solve using laplace transforms - it depends on where this is going, simple circuit theory (DC, ac and simple DC transient) wouldn't go past the answers above, if you're moving on to control systems or more advanced step functions then laplace is the way to go.
There are many different ways to solve the question posted - it's probably the way we have said, but it does depend on the level and type of course.
So I just asks for the current in one branch, so from ohms law I = V /Z, you know V, and you know Z as it's R+jX, X being 2 pi f L
This will be a complex number, so you can convert it to polar to get magnitude and phase angle.
Part II asks for the current in the other branch, so repeat the above remembering that capacitors lead and the calculation for Z is slightly different to an inductor.
Part III asks for the total current, so add the two branches, either via cartesian form as you have calculated above, or even in their polar form noting the sign /direction of the complex part.
Part IV is the impedance, so you could calculate it via standard series/parallel method (using the complex values), or as this has lead you down a simpler path Z=V/I - since you have V and the calculation of total current from part III - this would be an easy way to get the complex impedance
And so on...
I don't know how far you are down your course, but in any ac system calculations the use of complex numbers is pretty standard in anything but the simplest problem.
What level of course are you doing?
This is important as the more basic courses won't use complex numbers, and the more advanced would solve using laplace transforms - it depends on where this is going, simple circuit theory (DC, ac and simple DC transient) wouldn't go past the answers above, if you're moving on to control systems or more advanced step functions then laplace is the way to go.
There are many different ways to solve the question posted - it's probably the way we have said, but it does depend on the level and type of course.
"I really don't understand what I've done, but I guess I've done sonething?".
That has got to be the quote of the year
Good for you for giving it a go Bobby.
And we are only in January ? thanks happyhippydad"I really don't understand what I've done, but I guess I've done sonething?".
That has got to be the quote of the year
Good for you for giving it a go Bobby.
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This is what i have managed to do so far but i have a feeling i'm way off what is expected of me.
Can somebody tell me what J is? I know it was mentioned above that its the square root of -1 but when i type this into a calculator it says error
First thing...
You've used a mixture of values for frequency, 240Hz in one case, 50Hz in another. The SPICE parameters shown by the voltage source in the schematic show 50Hz but the question specifically states that the source is a variable frequency supply. I wonder what their intention was; If it is to provide a general solution for any frequency, then an f term will appear in every answer.
You've used a mixture of values for frequency, 240Hz in one case, 50Hz in another. The SPICE parameters shown by the voltage source in the schematic show 50Hz but the question specifically states that the source is a variable frequency supply. I wonder what their intention was; If it is to provide a general solution for any frequency, then an f term will appear in every answer.
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.
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.
Attachments
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?
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
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
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