Introduction to Electric Circuits

94 Single-phase a.c. circuits

The magnitude of I is thus u + Ic 2) and the phase angle 4~ is tan

(Ic/IR). Since I- V/Z, I R -- V/R and Ic = V/Xc we have

V/Z = V'[(V/R) 2 + (V//Xc) 2] = VV'[(1/R) 2 + (1/Xc) 2] - V~/(G 2 + Bc 2) = VY

In complex form we have

I(= V/Z) = IR + jlc = V/R + j V/Xc

so that

1/Z = 1/R + jl/Xc

and

Y = G + jBc

FigUre 4.33

(4.33)

Y

Bc

G

From the admittance triangle shown in Fig. 4.33 we see that the phase angle ~b is
given by tan -1 (Bc/G) = tan -1 (R/Xc) and since Xc = 1//toC, we have

th = tan-' (wCR) (4.34)

Example 4.20

A circuit consisting of a resistor of resistance 5 1"), in parallel with a capacitor of
10 IxF capacitance is fed from a 24 V, 4 kHz supply. Calculate the current in
magnitude and phase.

Solution

4kHz 24v19

I

El

5s _-- IOIJF

Figure 4.34

The circuit is shown in Fig. 4.34. From Equation (4.33) we have that the
admittance is Y = G + JBo Now

G = 1/R = 1/5 = 0.2 S