The resistance of a capacitor decreases The resistance of an inductor increases in .NET framework

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The resistance of a capacitor decreases The resistance of an inductor increases
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The Phasor Transform
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A phasor is a complex representation of a phase-shifted sine wave If f (t) = A cos( t + ) Then the phasor is given by F= A (732) (731)
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The Phasor Transform
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To see how this works, we begin by considering Euler s identity (714) Since e j( t+ ) = cos( t + ) + j sin( t + ), we can take (731) to be the real part of this expression, that is f (t) = Re[Ae j( t+ ) ] (733)
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If a source is given as a sine wave, we can always rewrite it as a cosine wave because cos(x 90 ) = sin x (734)
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In a given circuit with a source f (t) = A cos( t + ), the frequency part will be the same for all components in the circuit Hence, we can do our analysis by focusing on the phase lead or lag for each voltage and current in the circuit This is done by writing each voltage and current by using what is called a phasor transform We denote phasor transforms with boldface letters For f (t) = A cos( t + ), the phasor transform is just F = Ae j (735)
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The functions f (t) = A cos( t + ) and F = Ae j constitute a phasor transform pair We write this relationship as f (t) F EXAMPLE 7-2 If i(t) = 20 cos(12t + 30 ), what is the phasor transform SOLUTION First note that the current i(t) is the real part of i(t) = Re[20e j(12t+30 ) ] = Re[20 cos(12t + 30 ) + j20 sin(12t + 30 )] The phasor transform is I = 20e j30 Or we can write it in shorthand as I = 20 30
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(736)
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Circuit Analysis Demysti ed
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Properties of the Phasor Transform
Phasor transforms are very useful in electrical engineering because they allow us to convert differential equations into algebra In particular, the differentiation operation is converted into simple multiplication Again, let s start with f (t) = A cos( t + ), which allows us to write f (t) = Re [Ae j( t+ ) ] Then df = A sin( t + ) dt But notice that d Ae j( t+ ) = j Ae j( t+ ) dt Since f (t) = Re [Ae j( t+ ) ], it follows that df = j Re [Ae j( t+ ) ] dt For the phasor transform, we have the relation df j Ae j dt (738) (737)
Now let s consider integration We integrate from the time 0 just before the circuit is excited to some time t, so de ne g(t) =
t 0
Ae j( + ) d
Noting that at t = 0 , we take the function to be zero Letting u = j( + ) we have du = j d , d = 1 du j
And we obtain g(t) =
The Phasor Transform
1 j
Ae ju du =
1 Ae j( + ) j
t 0
1 Ae j( t+ ) j
Hence, integration, which is the inverse operation to differentiation, results in division by j in the phasor domain Given a sinusoidal function f (t) with phasor transform F we have the phasor transform pair
t 0
f ( ) d
1 F j
(739)
Circuit Analysis Using Phasors
With differentiation and integration turned into simple arithmetic we can do steady-state analysis of sinusoidally excited circuits quite easily The current owing through a capacitor is given by i(t) = C dv dt
When we work with phasors, this relation becomes I = j CV The voltage across a capacitor is related to the current via v(t) = 1 C
(740)
i( ) d
The phasor transform of this relation is V= 1 I j C (741)
Now let s turn to the inductor The voltage across an inductor is related to the current through the time derivative v(t) = L di dt
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