# RC/RL circuits and frequency selection

We have a lab that wants us to simulate an RC and RL circuit, and then a CR and LR circuit (series). From reading it is my understanding that reversing order of components results in reversing the action (ie frequency selection curve). In other words

LR=RC CR=RL

Why is this?

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If you draw the circuit diagrams and think about the voltage divider rule and how the component impedance magnitude changes with frequency, you'll have your answer. –  The Photon Apr 13 '12 at 20:39
Sorry, still not seeing it. If the resistor's frequency response is flat, how can its position affect the action? –  Joe Stavitsky Apr 13 '12 at 20:53
When you say "RC" circuit I thought you meant a resistor and a capacitor. When you say "RL" I thought you meant a resistor and an inductor. If that's not what you meant, could you explain those terms? –  The Photon Apr 13 '12 at 22:13
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## 2 Answers

There is a lot you implied or are assuming that you didn't state. The order of two components in series only matters if you access the node between them. It doesn't matter either if they are in parallel. You really should be more careful in explaining what you are talking about than just "RC and RL circuit". There are many ways to hook up resistors, capacitors, and inductors.

You apparently mean the special case of a RC and RL circuit where the two components are used as a frequency-dependent voltage divider. In that case the order matters just like it would if two different resistors are used in a voltage divider. Clearly these two circuits are not the same:

If the 10 kΩ resistors were replaced by capacitors, it would be even more obviously not the same. For the purpose of analyzing R-C filters, you can think of the C as a resistor that goes down with frequency. If R1 were replaced with a capacitor, the voltage divider would attenuate less at high frequencies and more at low frequencies, which we call a high pass filter. At DC it would attenuate infinitely, meaning it blocks DC. Similarly, if R3 were replaced by a capacitor, the voltage divider would attenuate more at high frquencies and less at low frequencies, which we call a low pass filter. At DC the capacitor looks like a open circuit, so it wouldn't attenuate at all at DC.

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$V_{OUT} = V_{IN} \cdot \dfrac{Z_C}{Z_C + R}$

If the impedance of $C$ is much higher than $R$ this equation will approach 1. This is the case for low frequencies, for DC the impedance will be infinite and then $V_{OUT}$ = $V_{IN}$.
For higher frequencies the factor $R$ will become more prominent, and if $R$ >> $Z_C$ the output will approach zero.

Now replace the resistor by an inductor, and the capacitor by a resistor. Then

$V_{OUT} = V_{IN} \cdot \dfrac{R}{R + Z_L}$

If $R$ is much higher than the impedance of $L$ this equation will approach 1. This is the case for low frequencies, for DC the impedance will be zero and then $V_{OUT}$ = $V_{IN}$.
For higher frequencies the factor $Z_L$ becomes more prominent, and if $Z_L$ >> $R$ the output will approach zero.

Note how similar the paragraphs after the equations are!

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