Hello,
this is from a previous exam and I do have a solution for it, but I do not understand it.
"Determine the current of the circuit in the time domain when \(Z_l\) is implemented as
inductance."
And the current source has been replaced with the following one:
\(
U_{in}(t) = 2*(sin(500*2{\pi}t) + sin(1000*2{\pi}t) + sin(1500*2{\pi}t))
\)
Okay, what I know:
Current in time domain should be something like this:
\(
i(t) = sin({\omega}t + {\psi}) \) (I have also seen \(\phi \) used instead of \(\psi\), am I confusing something here? The book I am using "Understandable Electronic Circuits" by Meizhong Wang uses \(\psi\)
[/tex]
From the previous task (this one here is 2.2, 2.1 uses the same circuit) I have:
\(
f_{lower} = 500 Hz
f_{higher} = 1500 Hz
\)
\(Z_l\) was supposed to have the value of 50 Ohms in order to get \(f_0 = 1 kHz\), I am unsure if I am supposed to re-use that here.
Anyway, onwards:
Looking at the voltage source, I notice the following:
\(
v(t) = V_{max} * sin({\omega}t)
\)
So I ASSUME, I do not know if this applies, please tell me, but I ASSUME \(V_{max} = 2V\).
Reasoning: I look at the voltage source. I see 3 sines and something "in front", the 2. If I use the definition above (taken from "Serway Physics for Scientists and Engineers", p.1035) then it has to be 2.
BUT and this confuses me, if I use, say, t = 7 and plug it in and use a calculator, I get 4.XX, which is greater than 2. So is \(V_{max} = 2V\) correct? I assume it has to be because of what follows, but how can I get 4.XX? This is probably very basic stuff but I really don't understand this. I am also unsure if I can get the maximum voltage that way, because I have 3 sines, instead of just 1.
Next:
\(
sin({\omega}t)
\)
\(\omega\) is defined as \(2*{\pi}*f\)
So if I look at the 3 sines, I see that the frequencies for those 3 are 500, 1000, and 15000.
And 500 is the lower cutoff frequency, and 1500 the higher cutoff frequency. What can I use this knowledge for?
Getting \(I_{max}\) I understand, \(V_{max}/Z = I_{max}\), and if I use
\(R = Z_l = Z_C \)
and then \(Z = sqrt(R^2 + (Z_C - Z_l)^2)\)
I get Z = 50 Ohms so \(I_{max} = 2/50\) (using the V_max from above), and according to my solution that's correct. So I guess I am allowed to use maximum voltage = 2V from above? It might be a trap or a crazy coincidence, but I am confused why I can get >2 from that equation.
This is as far as it made sense to me, assuming my logic is correct.
However, according to my solution, the following follows:
"The signal has 3 sinusoidal components which happen to be at the cut off frequencies and
the resonant frequency. Thus:"
\(
i(t) = (\frac{sqrt(2)}{2} * {I_{max}} sin(500*2{\pi}t + {\phi}_1) +{I_{max}} sin(1000*2{\pi}t) + \frac{sqrt(2)}{2} * {I_{max}} sin(1500*2{\pi}t +{\phi}_2)
\)
With \(I_{max} = \frac{2}{50} A\)
From Serway Physics p.1043:
\(
i = I_{max}sin({\omega}t - {\phi})
\)
(In RLC circuits)
This seems different from the solution, the \(\phi\) has a negative sign in the book, while in the solution it has a positive one? Is that because order of elements is different? In the book it is R-L-C in series, in the circuit here it is R-C-L. Why does that change the sign of the phi though?
Next part:
Current = Voltage / Resistance, in AC it is Current = Voltage / Impedance
I am going to skip a bit of the solution because I understand it, it is just simply expanding a fraction and simplyifing it, eventually it reaches this:
\(
I = {\frac{U_{max}(R - j({\omega}L - \frac{1}{{\omega}C}))}{R^2 +({\omega}L - \frac{1}{{\omega}C})^2}} \Rightarrow tan({\phi}({\omega})) = \frac{\frac{1}{{\omega}C} - {\omega}L}{R}
\)
How does that follow?
From the impedance triangle you get that phase angle between current and voltage is \(tan^(-1)(\frac{X_l -X_C}{R})\), but here \(X_l\) and \(X_C\) are swapped, why?
This has gotten quite lengthy so thanks for your time.
I am unable to use the "Upload a File" button, so I am using Imgur for the circuit instead:
http://i.imgur.com/Mcg7icW.png
this is from a previous exam and I do have a solution for it, but I do not understand it.
"Determine the current of the circuit in the time domain when \(Z_l\) is implemented as
inductance."
And the current source has been replaced with the following one:
\(
U_{in}(t) = 2*(sin(500*2{\pi}t) + sin(1000*2{\pi}t) + sin(1500*2{\pi}t))
\)
Okay, what I know:
Current in time domain should be something like this:
\(
i(t) = sin({\omega}t + {\psi}) \) (I have also seen \(\phi \) used instead of \(\psi\), am I confusing something here? The book I am using "Understandable Electronic Circuits" by Meizhong Wang uses \(\psi\)
[/tex]
From the previous task (this one here is 2.2, 2.1 uses the same circuit) I have:
\(
f_{lower} = 500 Hz
f_{higher} = 1500 Hz
\)
\(Z_l\) was supposed to have the value of 50 Ohms in order to get \(f_0 = 1 kHz\), I am unsure if I am supposed to re-use that here.
Anyway, onwards:
Looking at the voltage source, I notice the following:
\(
v(t) = V_{max} * sin({\omega}t)
\)
So I ASSUME, I do not know if this applies, please tell me, but I ASSUME \(V_{max} = 2V\).
Reasoning: I look at the voltage source. I see 3 sines and something "in front", the 2. If I use the definition above (taken from "Serway Physics for Scientists and Engineers", p.1035) then it has to be 2.
BUT and this confuses me, if I use, say, t = 7 and plug it in and use a calculator, I get 4.XX, which is greater than 2. So is \(V_{max} = 2V\) correct? I assume it has to be because of what follows, but how can I get 4.XX? This is probably very basic stuff but I really don't understand this. I am also unsure if I can get the maximum voltage that way, because I have 3 sines, instead of just 1.
Next:
\(
sin({\omega}t)
\)
\(\omega\) is defined as \(2*{\pi}*f\)
So if I look at the 3 sines, I see that the frequencies for those 3 are 500, 1000, and 15000.
And 500 is the lower cutoff frequency, and 1500 the higher cutoff frequency. What can I use this knowledge for?
Getting \(I_{max}\) I understand, \(V_{max}/Z = I_{max}\), and if I use
\(R = Z_l = Z_C \)
and then \(Z = sqrt(R^2 + (Z_C - Z_l)^2)\)
I get Z = 50 Ohms so \(I_{max} = 2/50\) (using the V_max from above), and according to my solution that's correct. So I guess I am allowed to use maximum voltage = 2V from above? It might be a trap or a crazy coincidence, but I am confused why I can get >2 from that equation.
This is as far as it made sense to me, assuming my logic is correct.
However, according to my solution, the following follows:
"The signal has 3 sinusoidal components which happen to be at the cut off frequencies and
the resonant frequency. Thus:"
\(
i(t) = (\frac{sqrt(2)}{2} * {I_{max}} sin(500*2{\pi}t + {\phi}_1) +{I_{max}} sin(1000*2{\pi}t) + \frac{sqrt(2)}{2} * {I_{max}} sin(1500*2{\pi}t +{\phi}_2)
\)
With \(I_{max} = \frac{2}{50} A\)
From Serway Physics p.1043:
\(
i = I_{max}sin({\omega}t - {\phi})
\)
(In RLC circuits)
This seems different from the solution, the \(\phi\) has a negative sign in the book, while in the solution it has a positive one? Is that because order of elements is different? In the book it is R-L-C in series, in the circuit here it is R-C-L. Why does that change the sign of the phi though?
Next part:
Current = Voltage / Resistance, in AC it is Current = Voltage / Impedance
I am going to skip a bit of the solution because I understand it, it is just simply expanding a fraction and simplyifing it, eventually it reaches this:
\(
I = {\frac{U_{max}(R - j({\omega}L - \frac{1}{{\omega}C}))}{R^2 +({\omega}L - \frac{1}{{\omega}C})^2}} \Rightarrow tan({\phi}({\omega})) = \frac{\frac{1}{{\omega}C} - {\omega}L}{R}
\)
How does that follow?
From the impedance triangle you get that phase angle between current and voltage is \(tan^(-1)(\frac{X_l -X_C}{R})\), but here \(X_l\) and \(X_C\) are swapped, why?
This has gotten quite lengthy so thanks for your time.
I am unable to use the "Upload a File" button, so I am using Imgur for the circuit instead:
http://i.imgur.com/Mcg7icW.png
