## Phasor Diagrams

We can understand AC circuits in terms of

phasor diagrams. Here a vector of length V_{o}rotates with angular velocity . Its projection onto the vertical axis isv = V _{o}sin tand that is just the AC voltage from a generator or a power supply.

The current i

_{R}through a resistor R is in phase with the voltage v_{R}across the resistor so we may show that on a phasor diagram by drawing a vector I_{R}which lines up with the voltage V_{R}.The current i

_{C}through a capacitor C is ahead of the voltage v_{C}across the capacitor by 90 degrees so we may show that on a phasor diagram by drawing a vector I_{C}which leads the voltage V_{C}by 90 degrees.The current i

_{L}through an inductor L is behind the voltage v_{L}across the inductor by 90^{o}so we may show that on a phasor diagram by drawing a vector I_{L}which lags the voltage V by 90 degrees.At any given time, the sum of the voltages across the three components, R, C, and L, must equal the voltage of the AC source. This is just energy conservation or Kirchoff's Loop Rule,

v(t) = v _{R}(t) + v_{C}(t) + v_{L}(t)or

v = v _{R}+ v_{C}+ v_{L}Remember that the phase relationship between the AC voltage v and each of the other voltages differs for each of the voltages. We can handle this graphically with the following phasor diagram:

or

For convenience, we may want to rotate this diagram,

Now we can see that the voltages form the sides of a right triangle with the voltage of the AC source forming the hypotenuse. That means

V = I SQRT[ R ^{2}+ (X_{L}- X_{C})^{2}]or, since V = I Z, we can now write the total impedence for this series RCL circuit as

Z = SQRT[ R ^{2}+ (X_{L}- X_{C})^{2}]where R is the resistance, X

_{C}is the capacitive reactance, and X_{L}is the inductive capacitance.[ Of course, the "SQRT" needs to be replaced with a properly written radical sign in the equations].

(c) Doug Davis, 2002; all rights reserved

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