Something Different:

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rattus said:
Ham, v(t) and i(t) are not constants; they are functions of time.

For the RC series circuit example, you could use any values of R and C you wish, e.g.,

R = 1/wC would provide a 45 degree leading current, but that would not be obvious if v(t) is a step function.
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In such a circuit, the capacitor should have a initial charge or just without charge.
 
drbond24 said:
... Perhaps useless would be a better word? ...
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Apparently my vote don't count, 'cause I said it was useless back in the early stages of this thread. And I meant in circuit analysis.

However, one thing plotting it out does serve... and that is to depict the 'opposition' to the flow of current as a function of time. For a purely reactive load, the tangent curve is centered about zero. When there exists both resistive and reactive components for the load, the curve skews positive.
 
Hameedulla-Ekhlas said:
In such a circuit, the capacitor should have a initial charge or just without charge.
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My guess would be with initial charge, or evaluate after circuit has stabilized. As I stated before, the power engineering aspect of this discussion usually considers the in-progress state of the event as the norm. Considering initial can and is done, but that is an atypical, "unless otherwise noted" condition.
 
Hameedulla-Ekhlas said:
rattus: Please see this example, is that what you mean

28aurlt.png
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Ham, looks right at first glance, but you should use RMS values instead of max values.

However, we haven't learned impedance and phasors yet, so why don't you solve the differential equation for i(t):

(The (t) is omitted to reduce the clutter)

v = iR + (1/C)int(i)dt + vco
 
Smart $ said:
My guess would be with initial charge, or evaluate after circuit has stabilized. As I stated before, the power engineering aspect of this discussion usually considers the in-progress state of the event as the norm. Considering initial can and is done, but that is an atypical, "unless otherwise noted" condition.
Click to expand...

I will reply you later first let rattus tell me capacitor has initial charge or not.
 
rattus said:
Ham, looks right at first glance, but you should use RMS values instead of max values.

However, we haven't learned impedance and phasors yet, so why don't you solve the differential equation for i(t):

(The (t) is omitted to reduce the clutter)

v = iR + (1/C)int(i)dt + vco
Click to expand...


I will expand this formula.
 
rattus said:
Ham, looks right at first glance, but you should use RMS values instead of max values.

However, we haven't learned impedance and phasors yet, so why don't you solve the differential equation for i(t):

(The (t) is omitted to reduce the clutter)

v = iR + (1/C)int(i)dt + vco
Click to expand...

Rattus:

Where did you get the vco in a mesh loop in above given equation

I have make it more clear the circuit see



28lan8n.png


1zc04ue.png



If I go through Laplace transform I can get the below equation easily

(R+1/Cs)= V(s)/I(s)
 
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Hameedulla-Ekhlas said:
Rattus:

Where did you get the vco in a mesh loop in above given equation

I have make it more clear the circuit see



28lan8n.png


1zc04ue.png



If I go through Laplace transform I can get the below equation easily

(R+1/Cs)= V(s)/I(s)
Click to expand...

Ham, vco is the initial voltage on the cap. You can let it be zero fo simplify the problem.

This problem is an example of the difficulty with solving instantaneous equations for such a simple circuit with differential equations as compared to the use of steady state methods where,

I = V/(R + 1/jwC)
 
Hameedulla-Ekhlas said:
rattus said:
Ham, vco is the initial voltage on the cap. You can let it be zero fo simplify the problem.

But rattus it must be zero otherwise the circuit will no more be linear.
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Ham, vco is an initial condition, that is all; it does not affect the linearity of the circuit.

For sinusoidal waveforms, I would let vco be zero. For a transient solution, vco might be of importance. You would account for it when you set the initial conditions.
Click to expand...
 
Smart $ said:
Apparently my vote don't count, 'cause I said it was useless back in the early stages of this thread. And I meant in circuit analysis.

However, one thing plotting it out does serve... and that is to depict the 'opposition' to the flow of current as a function of time. For a purely reactive load, the tangent curve is centered about zero. When there exists both resistive and reactive components for the load, the curve skews positive.
Click to expand...

OK, one vote for useless; now we have

"Meaningless, useless, nonsense"

As for the opposition to current, Ohm's Law is good enough. Students who have trouble grasping Ohm's law would be hopelessly confused by your Byzantine depiction. No cigar, not even a match.
 
rattus said:
As for the opposition to current, Ohm's Law is good enough. Students who have trouble grasping Ohm's law would be hopelessly confused by your Byzantine depiction. No cigar, not even a match.
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Yet the depiction is, in its bare essence, Ohm's Law.

Students studying to be an EE should seriously consider changing their major if they have a problem grasping Ohm's Law ;)
 
Hameedulla-Ekhlas said:
Smart$:
If the capacitor has an initial charge than this circuit will be no more a linear circuit.
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Still a linear circuit. If it has no initial charge, it will take a little operating time for the circuit to stabilize. How much time is dependent on circuit parameters.

An initial charge on the capacitor would be to mimic the stabilized operation from the get go.

In following example, both circuits are identical with the exception the second has an initial charge on the cap'. Graphs follow. Note the diffference is only within the first quarter cycle.

SomethingDifferent3.gif


SomethingDifferent4.gif
 
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