PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 239

v(t) I L sin t

I

C

m m sin( t)



() 

or

v(t) = XIm cos(t + θ)

where the total reactance is X = ωL − 1/(ωC).

Observe that the reactance X can be positive or negative. If X > 0 , then v(t) =

ImX cos(ωt + π/2) (inductive equivalent), and if X < 0 , then v(t) = ImX cos(ωt − π/2)

(capacitive equivalent).

R.3.41 Let us turn our attention to the LC parallel case.

Evaluate the current i(t) in the parallel LC circuit diagram shown in Figure 3.10,

assuming that the applied voltage is v(t) = Vmcos (ωt).

ANALYTICAL Solution

it()
 for BV cos tm   B

2

 0







where B represents the circuit admittance

it()
 for BV cos tm   B

2

  0

where B = BC^ − BL^ (adm it ta nce), BC^ =^ ωC, and BL^ =^ 1/(ωL).

R.3.42 Given a circuit where the current through and the voltage across are known. Then

let us evaluate the equivalent impedance, and its phasor diagram representation

for the following case:

i(t) = 2 cos(60 ⋅ 2 ⋅  ⋅ t − 30 °)

and

v(t) = 5 cos(60 ⋅ 2 ⋅  ⋅ t + 45 °)

FIGURE 3.9

LC series circuit diagram of R.3.40.

+ C –

i(t)

v(t)

+ L –

vL(t)

vC(t)

FIGURE 3.10

LC parallel circuit diagram of R.3.41.

i(t)

v(t) L C