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>> In this video, we're going to[br]demonstrate the actual process of

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analyzing circuits using[br]phasors and impedances,

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analyzing circuits in the phasor domain.

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We've already pointed out that in

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the phasor domain or rather[br]that we can analyze a circuit,

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in the phasor domain by representing[br]the voltages in terms of phasors,

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the currents in terms of phasors,

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and the impedances or representing[br]the effects of these capacitors,

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resistors, inductors in[br]terms of their impedances.

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So first of all,

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let's just represent phasor V,

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the source as having an amplitude[br]V sub m e to the j Theta.

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So, representing this time domain signal as

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its phasor with an amplitude of V[br]sub m and a phase shift of Theta.

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Now the impedance of the capacitor[br]is equal to one over

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j Omega C. The impedance of[br]a resistor is just equal to R,

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and the impedance of[br]the inductor is equal to j times

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Omega times L. So we'll[br]represent that or show that

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here is this impedance is one[br]over j Omega C. This is just R,

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and this impedance here is j Omega

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L. Now in the phasor domain

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there will be a current flowing[br]that we will represent as phasor I,

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and with these representations[br]let's go ahead now and write

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a Kirchhoff's voltage law equation[br]in terms of these phasers.

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Starting at this point and[br]going in this direction,

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I'm going to add up the voltage

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drops which means the voltage increase[br]will be a negative.

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So we'll have negative V sub m e

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to the j Theta plus now going in[br]the direction of current flow.

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So direct and current flow represents[br]a voltage drop and it will be a plus.

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The voltage drop across the capacitor is

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just the impedance of[br]the capacitor times the current

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or one over j Omega C times I,

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plus the voltage drop across the resistor[br]again going in the direction of

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current flow will be just R times I.

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Plus the voltage drop across[br]the inductor will be,

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the impedance of the inductor[br]which is j Omega L times I.

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The sum of those terms equals zero.

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Now, this is not a function of I.

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In fact, this is the constant[br]associated with the source.

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So let's bring it to the other side

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and factor out the I that is[br]common in each of those terms,

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we have in that I not equal[br]to I times one over j Omega C

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plus R plus j omega L is equal[br]to V sub m e to the j Theta.

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We now solve for I by[br]dividing both sides of

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the equation by this term in[br]parentheses, and we have then I,

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the phasor representation of[br]I is equal to V sub m e to

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the j Theta divided by one over j Omega C,

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plus R plus j Omega L.

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Phasor I the phasor representation is

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calculated by doing that complex division.