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By Ohm's Law, V = RI.

I am learning about impedance, reactance, etc.

One question: can Ohm's law be extended to impedance?

I mean if V = RI, will V = ZI?

Suppose this circuit:

enter image description here

Suppose this R/C represents an amplifier and I am injecting audio to it, from 20 to 20Khz.

The reactance of the capacitor will decrease with frequency. If AB represents the output of that circuit, what will happen to the voltage between A and B as the frequency increases from 20 to 20 kHz?

How to I plot that curve of Voltage AB with frequency?

Please answer these 2 questions. Thanks.

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    \$\begingroup\$ If 'AUDIO' represents an ideal voltage source, then nothing will happen to the voltage between A and B as frequency changes. The current flowing in the circuit will be a different question though .. \$\endgroup\$
    – brhans
    Jun 14, 2018 at 17:09

3 Answers 3

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Yes, Ohm's Law works with impedances but you must be careful to use a complex impedance. For example, in your circuit the total impedance is $$Z = R - \frac{j}{\omega C}$$ The resulting current will not be in phase with the applied voltage. More importantly, the voltage across the resistor and the voltage across the capacitor will be 90 degrees out of phase, so the magnitudes of these voltages will not sum to be equal to the magnitude of the source voltage.

To answer your specific question, if your ac source is ideal then the voltage from A to B will never change and is not a function of frequency. The voltage across an ideal source is not affected by whatever you connect to it, and the voltage from A to B is just the voltage across the source.

However, the impedance will vary with frequency so the current will vary with frequency. Therefore, the voltages across the resistor and capacitor will vary with frequency. It sounds like you want to create a Bode plot of these voltages.

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  • \$\begingroup\$ Thanks for the explanation. Just one question that I may have missed... why the negative signal on the capacitor reactance? \$\endgroup\$
    – Duck
    Jun 14, 2018 at 17:21
  • \$\begingroup\$ The negative sign for the capacitor's impedance (not reactance) is because the capacitor's impedance is -90 degrees phase angle from the resistor's impedance. An inductor's impedance is \$j\omega L\$ and has a +90 degree phase angle. \$\endgroup\$ Jun 14, 2018 at 17:27
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One question: can Ohm's law be extended to impedance?

I mean if V = RI, will V = ZI?

Yes, that's exactly why we use the concept of impedance.

The reactance of the capacitor will decrease with frequency. If AB represents the output of that circuit, what will happen to the voltage between A and B as the frequency increases from 20 to 20 kHz?

If the source is an (ideal-ish) voltage source, then the voltage between A and B will not change, but the magnitude of the current will increase with increasing frequency.

If the source is a current source (or a voltage source with nonzero internal impedance), then the magnitude of voltage will decrease with increasing frequency.

How to I plot that curve of Voltage AB with frequency?

Calculate the impedance of the RC combination at each frequency. Use the rules of complex math and your knowledge of the source's characteristics to determine the voltage at each frequency. Plot the result. You may want to plot phase as well as magnitude, or plot the response on a polar chart instead of a cartesian chart.

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Since \$Zc=\dfrac{1}{2\pi fC}\$ on a log-log scale it is a constant slope of impedance where 1uF passes through 100 Ohms at ~1.6KHz and it’s voltage is then 50% of the input.

Ideally voltage sources and Audio Amps are 0 Ohms so A and B are unaffected by the load.

But yes you can use Ohm’s Law at any one frequency using Xc(f).

Extra info

But the current will follow the apparent impedance of the square root of two Cartesian vector values squared to get the hypotenuse or magnitude of impedance the angle determines the phase of current relative to voltage .

For a graphical view of RLC impedance see the chart on another of my answers .

How to calculate the gain of a RLC tank filter?

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