AC Circuit analysis

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#1
Guys can anyone help me with this question please??
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7 years ago
#2
(Original post by ChemBoss)
Guys can anyone help me with this question please??

What are you stuck on? What don't you understand?
Have you done the theory and know what a phasor diagram is?
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7 years ago
#3
(Original post by ChemBoss)
Yeah I understand the theory but I'm not 100% certain about it. From my understanding you would have to find the phasors of the voltage and divide it by the impedances of the resistor and inductor. I know the impedance for the resistor is just R and for the inductor its jwL. What I don't understand is the concept of phase diagrams and how Im supposed to add the two impedances together in the Euler notation.

I would appreciate it if you clarify this for me.
For a simple calculation like this I would
Start by finding the total impedance of the circuit. You know R and XL.
You add them using Then find current using V=IZ
The pd across the inductor leads the pd across the resistor (and the current in the circuit) by 90 degs.
The standard formula for phase angle is This shows you how the phasor diagram looks.
Diagram on 2nd page

2
#4
Ahhhh okay can I just ask though how did you know that the inductor leads the pd across the resistor by 90 degrees?

Also, obviously this is quite a simple example, is there a textbook step by step way of working these types of questions out for more complex questions?
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7 years ago
#5
(Original post by ChemBoss)
Ahhhh okay can I just ask though how did you know that the inductor leads the pd across the resistor by 90 degrees?

Also, obviously this is quite a simple example, is there a textbook step by step way of working these types of questions out for more complex questions?
In ac theory one simply states that the pd across a pure inductance leads the current by 90 degs. Similarly, as you see in the scanned attachment, you state that the pd across a pure capacitance lags behind the current by 90 degs. This is a standard assumption in these circuits. These two assumptions enable you to draw the phasor diagram and lead to the formulas I've mentioned.
The scans I attached explain this. It's worth reading through them.
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#6
Ahhh okay thank you ever so much. Can you please tell me what book you extracted these notes from?
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7 years ago
#7
(Original post by ChemBoss)
Ahhh okay thank you ever so much. Can you please tell me what book you extracted these notes from?
It's book I've used many times here. An old A-Level text from the 1980s. Tom Duncan - Physics - A textbook for Advanced Level Students.
This edition would be considered out of date for modern specifications but is much more in depth for the more traditional topics.
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#8
(Original post by Stonebridge)
It's book I've used many times here. An old A-Level text from the 1980s. Tom Duncan - Physics - A textbook for Advanced Level Students.
This edition would be considered out of date for modern specifications but is much more in depth for the more traditional topics.

These notes are really helpful man Ive learnt so much from them. Can you please upload the next couple of pages if it shows you how to deal with parallel curcuits please?

I have this question here which I feel like I can do if I can add the 15nF and the 6.8nF capacitors in parallel, am I allowed to do that? If I am then would I just find the total impedance of the curcuit by using 1/Ztot=1/jwc1+1/jwc2 and then calculate Phasor of voltage/Total impedance?

Honestly thank you so much for your help it really has helped a lot.
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7 years ago
#9
Yes you can add the capacitors that are in parallel to produce one equivalent capacitance.
You can also then add that to the other (10nF), in series, to find the total capacitance for the circuit.
From this you can find the total impedance of the circuit. And hence the current.
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#10
(Original post by Stonebridge)
Yes you can add the capacitors that are in parallel to produce one equivalent capacitance.
You can also then add that to the other (10nF), in series, to find the total capacitance for the circuit.
From this you can find the total impedance of the circuit. And hence the current.
So just to clarify there is only one current phasor throughout the circuit? If so, the question saying phasors is just a mistake?

So just to summarise,
1. Add all the currents to find a capacitor of capacitance 31.8
2. The impedance of this is then equal to -j(2pi(1000))(31.8x10-9)
3. We then do 100√2 divide by the above impedance to find the current phasor?

Is this correct?
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#11
(Original post by Stonebridge)
Yes you can add the capacitors that are in parallel to produce one equivalent capacitance.
You can also then add that to the other (10nF), in series, to find the total capacitance for the circuit.
From this you can find the total impedance of the circuit. And hence the current.
oh sorry no the total capacitance would be 1/(1/15.8)+(1/10))...and then the same there on in?
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7 years ago
#12
(Original post by ChemBoss)
oh sorry no the total capacitance would be 1/(1/15.8)+(1/10))...and then the same there on in?
Yes.
The current found from this (when converted to impedance) is the rms current in the main circuit, delivered by the source.
The pd across the 10nF capacitor plus the pd across the parallel capacitors adds to give 10Vrms So you have two different pds here.
Also the current in the two parallel capacitors is not the same. They each have an amount that adds to give the current in the main circuit. (It's not unlike working with resistors.)
These things need to be calculated.
As you have pure capacitance and no resistance you know the phase angle between the pd across a capacitor and the applied pd.
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#13
(Original post by Stonebridge)
Yes.
The current found from this (when converted to impedance) is the rms current in the main circuit, delivered by the source.
The pd across the 10nF capacitor plus the pd across the parallel capacitors adds to give 10Vrms So you have two different pds here.
Also the current in the two parallel capacitors is not the same. They each have an amount that adds to give the current in the main circuit. (It's not unlike working with resistors.)
These things need to be calculated.
As you have pure capacitance and no resistance you know the phase angle between the pd across a capacitor and the applied pd.
Ahhh okay thats make sense, so does that therefore mean to find the way the 10Vrms is split up you have to use Q=CV? Basically acting as if the circuit is DC? And also is that phase angle 90 that you're talking about?

Also as the V value is Vrms it is therefore equivalent to the DC V value and we can use it as that or is that incorrect?
0
7 years ago
#14
(Original post by ChemBoss)
Ahhh okay thats make sense, so does that therefore mean to find the way the 10Vrms is split up you have to use Q=CV? Basically acting as if the circuit is DC? And also is that phase angle 90 that you're talking about?

Also as the V value is Vrms it is therefore equivalent to the DC V value and we can use it as that or is that incorrect?
To find Vrms for, say, the 10nF capacitor you use Vrms = IrmsZ where Z is its impedance.
You found Irms earlier.
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#15
(Original post by Stonebridge)
To find Vrms for, say, the 10nF capacitor you use Vrms = IrmsZ where Z is its impedance.
You found Irms earlier.
Where do we know Irms from?

Besides do we not need the peak Voltage value for Vsin(wt) and not the Vrms value?
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7 years ago
#16
(Original post by ChemBoss)
Where do we know Irms from?

Besides do we not need the peak Voltage value for Vsin(wt) and not the Vrms value?
I explained how you get Irms in the main circuit in post #12.

Why do you need the peak value?
If the question gives you rms value of voltage, then use it and find the rms current.
If ever you need the peak value just multiply the rms value by root2
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