Introduction to Electric Circuits

Vc = IXc ~l VL = IXL

~~] I (reference)

VR = IR

(a)

VL = IXL

0

Vc = IXc

Figure 4.20

VR = IR

r

(c)

4.3 Series a.c. circuits 81

VL =IXL

/

v. = IR/

V

I (reference)

r

Vc = IXc

r

I (reference)

and

V- I~/(R 2 + [X L - Xc] 2)

The impedance is

Z (-V/I) - V(R 2 "}- [X L - Xc] 2)

The phase angle is obtained from cos 6- R/Z or sin 6- (XL- Xc)/Z or
tan 05 = (XL- Xc)/R.
In the second case with Xc > XL we see that the current leads the voltage V
by the phase angle 05 and that therefore the circuit behaves as a capacitive
circuit. Also we have that V 2 = (IR) 2 + (I[Xc - XL]) 2 and so

V = I%/(R 2 + [X c --XL] 2) (4.18)

The impedance is Z=x/(R2+[Xc--XL] 2) and the phase angle is
~b = tan -1 (Xc- XL)/R.
There is a third possibility for this circuit: that the inductive reactance XL is
equal to the capacitive reactance Xc. In this case the phasor diagram takes the
form shown in Fig. 4.20(c), from which we see that V = IR, confirmed by
putting Xt. = Xc in Equation (4.18). The circuit then behaves as a purely
resistive circuit and is a special case which is fully discussed in Chapter 6.

Example 4.9

In the circuit of Fig. 4.19, R - 12 1), L - 150 mH, C - 10 I~F, v = 100 sin 2"rrft
and f= 100 Hz. Calculate (1) the impedance of the circuit, (2) the current
drawn from the supply, (3) the phase angle of the circuit. Give an expression
from which the current at any instant could be determined.