0195136047.pdf

3.1 SINUSOIDAL STEADY-STATE PHASOR ANALYSIS 115

EXAMPLE 3.1.5

Consider anRLCseries circuit excited byv(t)=

(

100

√

2 cos 10t

)

V, withR= 20 ,L= 1

H, andC= 0 .1 F. Use the phasor method to find the steady-state response current in the circuit.
Sketch the corresponding phasor diagram showing the circuit-element voltages and current. Also
draw the power triangle of the load.

Solution

The time-domain circuit and the corresponding frequency-domain circuit are shown in Figure
E3.1.5. The KVL equation is

V ̄=V ̄R+V ̄L+V ̄C=I ̄

(

R+jωL+

1

jωC

)

Note thatω=10 rad/s in our example, and rms values are chosen for the phasor magnitudes.
Thus,

100 0°=I ̄( 20 +j 10 −j 1 ) or I ̄=

100  0°

20 +j 9

=

100  0°

21. 93  24 .23°

= 4. 56  − 24 .23° A

Theni(t)=Re

[√

2 Ie ̄ jωt

]

, which is

i(t)=

√

2 ( 4. 56 )cos( 10 t− 24 .23°)A.

The phasor diagram is shown in Figure E3.1.5(c).

−

+−+

+

−

−

v(t) = 100 2 cos ωtC = 0.1 F V =

vR vL +−+

+

=

−

−

VR VL

vC VC

i(t)

R = 20 Ω L^ = 1 H jωL^ =^ j^10 Ω

Ω

R = 20 Ω

(a)

−

+

100 0°V I

(b)

jωC

1

j

(^1) =−j
φ = 24.23
VL = 45.6∠(90 − 24.23) = jXL I
(VL + VC)
VR + VL + VC = V = 100 ∠ 0 V
VC = 4.5 ∠−(90 + 24.23)
= jXC I
VR = 91.2 ∠−24.23 = IR
(c)
I
Figure E3.1.5(a)Time-domain circuit.(b)Frequency-domain circuit.(c)Phasor diagram.(d)Power
triangle.