Hameedulla-Ekhlas
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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.
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.
rattus: Please see this example, is that what you mean
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
Hameedulla-Ekhlas
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I will reply you later first let rattus tell me capacitor has initial charge or not.
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Hameedulla-Ekhlas
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Rattus:
Where did you get the vco in a mesh loop in above given equation
I have make it more clear the circuit see
If I go through Laplace transform I can get the below equation easily
(R+1/Cs)= V(s)/I(s)
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Rattus:
Where did you get the vco in a mesh loop in above given equation
I have make it more clear the circuit see
If I go through Laplace transform I can get the below equation easily
(R+1/Cs)= V(s)/I(s)
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
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Hameedulla-Ekhlas
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Smart$:
If the capacitor has an initial charge than this circuit will be no more a linear circuit.
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.
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
This is the nonsense part.
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.
I don't know how you can say that.
Is not v(t) equal to E at each instant.
Is not i(t) equal to I at each instant.
v(t)/i(t) = E/I = R ...at each instant.
No, no, maybe.
Please explain...
Last time I checked :roll:, volts is volts and current (i) is current... and this is what E and I symbolically represent, respectively.
E and I represent RMS values, e and i represent instantaneous values.
But the RMS values of an instant are e and i.
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