LCR circuits

Kevin Weddle

Feb 23, 2004
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Has anybody seen the calculations for the more complicated LCR circuits. I understand the basic circuits, but any addition is not well illustrated in the book I have. Take a simple series LC and parallel the capacitor with a resistor. This goes beyond what I have seen being calculated. If a circuit is nowhere near the resonant frequency, can we assume the regular impedance of the capacitor and inductor?

We could start with voltage subtraction. The inductive reactance tends to subtract from the capacitive reactance. A parallel resistor will tend to make it an LR or RC circuit
XL = 50
XC = 100
then impedance is 50
Make it an RC circuit R=200
then impedance = 300
so the impedance is somewhere between these two values.

 
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russlk2

Feb 8, 2004
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You can't use the reactance as a plain number, you have to consider the phase angel. The current leads the voltage by 90 degrees in a capacitor and lags by 90 degrees in an inductor. The reactance of an inductor is given by Xl = 2*PI*f*L
where PI=3.14162, f= frequency in Hz, L=inductance in henries.
The reactance of a capacitor is given by Xc = 1/(2*PI*f*C).
The quantity 2*PI*f is given the symbol ohmega, but since I don't have that symbol on the keyboard, I will use W. The 90 degree phase angle is denoted by j, so now Xl = j*W*L and Xc = 1/(j*W*C).

The advantage of this nominclature is that it is easy to add series impedances: Z = Xl+R = j*W*L+R
Parallel impedances is the familiar product over sum: Z = j*W*L*R/(j*W*L+R).
You can convert a parallel circuit to an equivalent series circuit and vice-versa. I used to be able to practically do that in my head, but it has been many years.

 

Kevin Weddle

Feb 23, 2004
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I use vectors any way I can. Formulas are easy to work with but, in the case of electronics, they don't make much sense. We need a way to make sense of the relationship without turning it into a sequence of mathematical operations. Do you understand what I mean?

 
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