ByLisJo
Can someone please help me with b and c?
I don't really have a clue about the b, but the current flowing from A to B should be equal to the current flowing from B to C. And the total current of a loop (e.g., the loop between A and B) is calculated by adding the current through R and the current through the capacitance? Can this do anything?
I have absolutely no approach to the c
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Can someone explain to me how this works?
In physics we are currently doing pulleys and I don't understand anything Thank you in advance for your answers
LG H.
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LG H.
Thank you for your detailed answer. I do not quite understand how you come to the -1 at c) and then say that wCR is equal to one
At a) you add the impedance of AB and BC.
Yeah.
Yes, but added geometrically or complexly.
At b) and c), you take your equation for the impedance you have determined in a). There you use the term CR^2 for L. Tip: Set in the first term of the equation (for Z_AB) preferably C = L/(R^2). You just leave the second term as he is and add everything.
Spoiler: The complete equation will only be reduced to Z_AC=R. In this way, you have phase-equality with the source voltage in the current. The circuit behaves like an LC series resonant circuit. The current through C is forwarding the voltage through 90°, while the current through L is continuing through 90°. Due to L=CR^2, both shares are equal. As a result, they lift up and the pure active component remains.
Thank you. a and b I understood now, but I still don’t understand the c. Maybe I’m standing on the squeegee, but I don’t understand why L=CR^2 makes the naparts hleich big. And what exactly is meant by proportion and pure proportion of action?
And don’t you calculate the phase difference as to the arctan of Real by imaginary part?
The “resistance” of a circuit with inductances and/or capacitances is called impedance Z, namely because the resistors do not have purely ohmic, but also complex or blind portions.
The ohmic portion is described with the real part, also called active part R. The inductive/capacitive portion is described with the imaginary portion, also called reactive portion X.
Together, they give the impedance Z=real part R+imaginary part X
With a), we first determined the equation for the total impedance Z_AC of the circuit.
(b) we have used the term L=CR^2 and found that we will receive Z_AC = R at the end. In other words, the impedance here consists only of a real part.
This in turn means that the imaginary part or Blind component Totalimpedance Z_AC is no longer present. -> X_AC = 0
In other words, the imaginary part of Z_AB+ must yield the imaginary part of Z_BC equal to 0.
That both shares “lift” up.
Look at @Halseddystrom’s answer! It’s very detailed.