Introduction to Electric Circuits

80 Single-phase a.c. circuits

As in the case of the RL circuit we can divide each of the phasors in
Fig. 4.18(a) by I to obtain the impedance triangle for the circuit and this is
shown in Fig. a.18(b). We see that cos ch=R//Z; sin ch=Xc//Z and
tan ~b - Xc//R where ~b is the phase angle of the circuit.

Example 4.8

In the circuit of Fig. 4.17, R = 10 f~, C = 10 IxF and f= 400 Hz. Calculate (1)
the impedance of the circuit, (2) the phase angle of the circuit.

Solution

1 The impedance Z = ~v/(R 2 + Xc2).

Now R = 10 f~ and Xc = 1/2.afC- 1/(2zr 400 x 10 • 10 -6) = 39.8 fl, so

Z = ~/(102 + 39.82) = 41 1)

2 The phase angle of the circuit is
4~- tan-1 (Xc/R) = tan -~ (39.8/10) = 75.9 ~

Series RLC circuits

__~ R L

vtO

Figure 4.19

<VR ~VL

c

II

"~ Vc

Applying KVL to the circuit of Fig. 4.19 and taking the clockwise direction to

be positive,

V- VR- VL- Vc-0 (phasorially)

V= VR + VL + Vc- IR + IXL + IXc (phasorially)

The phasor diagram is drawn in Fig. 4.20(a) assuming that XL > Xc and in

Fig. 4.20(b) assuming that XL < Xc. In both cases the voltage VR (=IR) is

drawn in phase with the reference phasor (I), the voltage VL (=IXL) is drawn

90 ~ ahead of the current I and the voltage Vc is drawn 90 ~ behind the current I

in accordance with Equations (4.12), (4.13) and (4.14), respectively.

In the first case (Fig. 4.20(a)) we see that the current lags the voltage V by the

phase angle 4' and that therefore the circuit behaves as an inductive circuit.

Also we have that

v (m) + xo]