Introduction to Electric Circuits

96 Single-phase a.c. circuits

12 --" (IL COS (])L) 2 q- (/L sin q~c -- Ic) 2 (4.35)

and

~b = tan -1 [(Ic sin qSC -- IC)/IL COS qSL (4.36)

The case where q5 turns out to be 0, i.e. when Ic sin qSC = IC is a special case

which will be fully discussed in Chapter 6.

In complex form, we have, from the phasor diagram of Fig. 4.35(b),

I = Ic COS qSC + j(Ic -- Ic sin ~bL) (4.37)

If Ic > Ic sin qSC then the j term is positive indicating that the total current

leads the voltage and that the circuit is therefore predominantly capacitive.

If Ic < Ic sin 4K then the j term is negative indicating that the total current

lags the voltage and that the circuit is therefore predominantly inductive.

Also we have that V = Vc = VR + VL phasorially.

Example 4.21

Determine (1) the total current drawn from the supply in the circuit of Fig. 4.36,

(2) the phase angle of the inductive branch and of the circuit as a whole.

IOOV 0

I

[0 5~ ~_20fl

l,~176

Figure 4.36

Solution

1 The current through the capacitor is Ic - V/Xc = 100/20 = 5 A. The

current through the inductive branch is IL = V/ZL where ZL -- R + jXL. The

phase angle of the inductive branch is given by

4~L- tan -~ (XL/R) = tan -1 (10/5) - 63.4 ~

ZL = V(R 2 + XL 2) --" V(5 2 Jr- 10 2) = 11.18 a

Ic- V/ZL- 100/11.18- 8.9 A

From Equation (4.35)

12 -- (Ic COS 4L) 2 + (IL sin qSc -- Ic) 2

= (8.9 COS 63.4) 2 + (8.9 sin 63.4 -- 5) 2 = 15.9 + 8.75 = 24.65 A 2

Therefore

I= 4.96 A