PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

234 Practical MATLAB® Applications for Engineers

R.3.29 Let us assume that the voltage across the capacitor C is vC(t) = Vm cos(ωt) in volts,
then its current is given by

it C

dv t

dt

C

d

dt

C() c Vtm

()

[cos()]

itCm()
CV sin( )t
 CVmcos t



2









Let

it ICm()
cost



2









then

ICVmm


or by Ohm’s law

X

jC

C() j







1

2

where denotes a phase angle of







R.3.30 Let the voltage across C be vC(t) = Vm cos(ωt) (from R.3.29), then the current through
C is given by iC(t) = Imcos(ωt + (π/2)). Clearly, the current leads the voltage by an
angle of π/2 rad. Capacitive reactance XC represents the opposition to the fl ow of
charge, which results in the continuous interchange of energy between the source
and the electric fi eld of the capacitor.

R.3.31 Let the current be given by i(t) = Im cos(ωt), in the series RL circuit shown in Fig-
ure 3.1. Then,

vR(t) = RIm cos(t)

vL(t) = −L Im sin(t) (from R.3.26) and applying KVL,

v(t) = vR(t) + vL(t)

v(t) = RIm cos(t) − LIm sin(t)

vt()
Im R^22 ( ) cos(L t)





   tan^10

2

L

R









for

FIGURE 3.1
RL circuit diagram of R.3.31.

+ L –

+ R –

vR(t) i(t)

v(t)

vL(t)

+ –