Question: difference between RC, RL, LC

How do you tell the difference between RC, RL, LC circuits?

Each of these circuits, and what they qualitatively do in DC and AC circuits...

RC - a resistor and capacitor in series. Exhibits charging behaviour with characterisitic time constant with DC voltage source. Acts as a high pass filter (allows high frequency currents, but not low frequency currents) in AC circuits.
RL - a resistor and inductor in series. Acts as a short with a DC voltage source, but smooths out rapid variations in current. Acts as a low pass filter (allows low frequency currents, but not high frequency currents) in AC circuits.
LC (and RLC) - an inductor and capacitor (and resistor) in series. If initially charged, has oscillitory behaviour (damped if also has a resistor). Has resonant behaviour with AC driving voltage (damped if also has a resistor).

Question: LC in parallel

How to calculate the resistance of a capacitor and an inductor connected in parallel?
(-XcXL/(XL-Xc))^2 ?

This is a little bit trickier that considering elements in series because the inverse of the impedances have to be added. I'm pretty sure it is beyond the scope of this course to need to do this calculation, but I will put it here anyway. I'm also not sure how you would arrive at the answer without using the appropriate complex represenations of the impedance, so I will use them but try to be careful to explain.

So, for elements in parallel:

1/Z_tot=1/Z_C+ 1/Z_L+ 1/Z_R ...[1]

We will consider just an LC circuit (with ideal components), so R=0. Now, in complex notation Z_Cand Z_L are:

Z_C=1/iωC ...[2]
Z_L=iωL ...[3]

So substituting [2,3] into [1]:

1/Z_tot=iωC+1/iωL

or, using the fact that 1/i=-i:

1/Z_tot=iωC-i/ωL
1/Z_tot=i(ωC-1/ωL) ...[4]

so, we can now replace 1/ωC with X_C and ωL with X_L:

1/Z_tot=i(1/X_C-1/X_L) ...[5]

Now we can simplify the part in the brackets, giving us:

1/Z_tot=i(X_L-X_C)/X_CX_L ...[6]

or taking the inverse of the fraction on both sides (and once again using 1/i=-i):

Z_tot=-iX_CX_L/(X_L-X_C) ...[7]

Now, if we had also had a resistor in parallel in the circuit (R), we would end up at equation [7] with both real and imaginary parts. To find X_tot we would have to find the magnitude of the resultant phasor. Since Z_tot is purely imaginary, we need only to find the magnitude of this number, which we can express as:

X_tot=-X_CX_L/(X_L-X_C) ...[8]

Which is pretty much what I think you suggested. Note that if we hadn't used the imaginary numbers we would not have been able to get the right signs and would have come up with a different answer.

Looking back at eqn. [7], let's think about what this represents... In the imaginary plane Z_tot would be a phasor pointing along the y-axis. For values of X_L and X_C which make Z_tot positive, the impedance could be considered "inductive", and inversely if Z_tot were negative, the impedance could be considered "capacitive".

Let's consider two extreme cases...

If the frequency is high, X_L is very large, and X_C is very small (let's say negligible). In the numerator of Z_tot the frequency cancels out, so this is a constant. If we look at the denomenator, X_L >> X_C, so the denomenator is approximately X_L. Since X_L is large, Z_tot will be small and negative (capacitive). In fact if you substitute in equations [2,3] ignoring X_C in the denominator, you will recover Z_C. This is consistant with the inductor providing a break in the circuit, and the capacitor providing a short.

Similarly, if the frequency is very low, X_C becomes very large, and X_L very small. In this case, the result is small and positive (inductive), and indeed again, if you work it out you will recover Z_L as Z_tot. This is consistant with the capacitor providing a break in the circuit, and the inductor providing a short.

To summarize:

ω→∞ Z_tot→Z_C and current flows through the capacitor
ω→0 Z_tot→Z_L and current flows through the inductor

These limiting cases can be worked out simply as a thought experiment (consider which branch of the circuit has a small resistance and which has a large resistance) and are often more insightful than getting through the algebra.

Hope this helps.

Topic Open: RC, RL, LC, LRC circuits

The following topic is now open for questions: RC, RL, LC, LRC circuits (and any AC circuits)

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Question: Impedance of LR circuit with "real" inductor

Just had a question about LR circuits. Just say there was a circuit consisting of a resistor and inductor. And just say the inductor had a resistance of its own. How do you find the total impedance of the circuit. This was similar to one of the CAPA questions. My friend told me to find the impedance of just the resistor and inductor and then add the resistance of the inductor to Z. I am confused on why you do this? Why couldn't you just add the resistance of the inductor and resistor first, then use the formula Z = sqrt(R^2 + XL ^2), R being the resistance of the inductor and resistor combined?

Actually, you are right! You can (and should) just add the resistance of the inductor to the resistance of the resistor then find the impedance including the inductive term.

Here's how it works. A real inductor will have some inherent resistance due to the fact that it is a (usually fairly large) coil of wire. To analyze a circuit, we would replace a "real" inductor with and "ideal" inductor and a resistor (to represent the pure resistance of the inductor). So, in an LR circuit you would go from having 1 resistor and 1 "real" inductor in series to having 2 resistors in series and one "ideal" inductor also in series. You would then find the net resistance of the circuit, and this would be "R" in the formula you quote above.

From this you would find the magnitude of the inductance from Z=sqrt(R²+X_L²)
(this formula comes from the fact that R and X_L are 90^o out of phase, so the magnitude is the hypotenuse of the two "phasors" see: Basics of AC circuits/RLC circuit example). Since R and X_L do not add linearly, I do not think you should get the right answer if you go about it in the opposite order (except maybe in some special circumstance).

EDIT: A similar question was also asked regarding an LRC circuit. In this case, the methodology is the same. The inductor with a resistance is replaced by a resistor and an inductor in series. The resistance of the whole cicuit is calculated, and then the magnitude of the impedance is determined. In the case of an LRC circuit, Z is given by:

Z=sqrt(R²+X_L²+X_C²)

Otherwise, the whole problem is the same.

Hope this helps!