0:00:00.000,0:00:03.150
>> We're now going to derive[br]the Impedance of the Capacitor.

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Again impedance we have defined as

0:00:06.150,0:00:09.660
being the ratio of the phasor[br]voltage to the phasor current.

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So let's determine that relationship[br]now for our capacitor.

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Once again, we're going to define[br]a voltage or reference voltage v of

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t and a current flowing through[br]the capacitor flowing or as reference,

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from the positive to[br]the negative reference to terminal.

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Now this time around, let's[br]assume that the voltage and

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the capacitor is equal to[br]or is of the form V sub m,

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cosine of Omega t plus Theta sub V.

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We know that in a capacitor[br]i is equal to c times

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the derivative of the voltage,

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which is going to equal[br]then c. Now once again,

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the derivative of cosine[br]is negative the sine,

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so we'll bring that[br]negative out in front here,

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the chain rule is going to[br]give us an Omega out in front.

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We have then the V sub m and[br]we will have in the sine,

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the derivative of cosine is the sine of the

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negative already taken[br]care of sine of Omega t

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plus Theta sub V. Once again,

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we want to write this in terms[br]of cosine rather than a sine.

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So this is going to equal negative C times

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Omega times V sub m times the cosine

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of Omega t plus Theta[br]sub V minus 90 degrees.

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Now, let's write both of[br]these and their phasor forms.

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This time we have phasor V is equal to,

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V sub m e to the j Theta sub V.[br]We can write phasor I is equal to

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negative negative C Omega V sub m, e

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to the j Theta sub V minus 90 degrees.

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Again, using the product of

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two exponential terms is that you[br]add the exponents and do that.

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So we have the answer is equal[br]to negative C Omega V sub m e to

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the j Theta sub V e to[br]the minus j90 degrees.

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As before, this term right[br]there is just minus j.

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So we take the minus times[br]a minus that gives us a positive,

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and bringing the J into this makes[br]here we have then J Omega C,

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V sub m e to the j Theta sub V. Again,

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we recognize that right there as being

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phasor V. So we can then write that phasor

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I is equal to j Omega[br]C times phasor V. Now,

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let's go ahead and form the impedance by

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writing a ratio of phasor V to phasor I.

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We have Z then is equal to[br]phasor V divided by phasor I,

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but phasor I is just j Omega C times[br]phasor V. The phasor V is cancel.

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We get then that the impedance of[br]a capacitor is equal to one over

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j Omega C. Sometimes we'll write that by[br]bringing the j up into the numerator,

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changing the sign on it which then gives us

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a negative J times 1[br]over Omega C. All right.

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We need to make some observations here.

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First of all, notice that the impedance[br]of the capacitor is negative.

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Second of all, notice

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the frequency dependency of[br]the capacitor impedance is inversely.

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In other words, I'll say it better,

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the impedance of a capacitor[br]is inversely proportional,

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to the frequency of[br]the circuit source that's driving it.

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Remind ourselves here.

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The impedance of an inductor[br]was j times Omega

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L. The impedance of an inductor was[br]directly proportional to the frequency.

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The impedance of a capacitor is[br]inversely proportional to the frequency.

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When the frequency is small,

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the impedance of the capacitor[br]is large and in fact,

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that DC when Omega equals 0,

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the capacitor has an infinite impedance.

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It is effectively an open circuit.

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As the frequency gets larger and larger[br]as the frequency approaches infinity,

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a larger number in the denominator makes[br]the impedance of the capacitor gets

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smaller and smaller and in the limit[br]as Omega goes to infinity,

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the impedance capacitor goes to zero.

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At high frequencies, the capacitor[br]is approximated by a short circuit.

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So for example, let's say that we've[br]got a 10 microfarad capacitor.

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The source is driving this is some V sub

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s of t is equal to 5 cosine of 5,000t.

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In other words, Omega is equal[br]to 5,000 radians per second.

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Then the impedance of[br]the capacitor in this circuit,

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this 10 microfarad capacitor would be[br]equal to 1 over j times Omega which is

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5,000 times c which is 10[br]times 10 to the minus 6.

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You do the math on that, again[br]bringing the j in the numerator.

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Here you'll get that the impedance[br]of this capacitor would

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be negative negative j20.

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As we mentioned, the impedance[br]of the capacitor is negative,

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the impedance of the inductor is positive.

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The impedance of the inductor,

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in fact, let's just make[br]a little chart here.

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In the time domain R and the impedance for

0:06:05.390,0:06:10.400
a resistor is just plain[br]old R. For a capacitor,

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the impedance of a capacitor[br]is 1 over j Omega C.

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For an inductor, the impedance[br]of the inductor is J Omega L.

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Finally, in each of[br]these cases, Ohm's law applies.

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We have then that V is equal[br]to I times Z because we

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defined Z to be the ratio[br]of phasor V to phasor I.