Introduction to Electric Circuits

78 Single-phase a.c. circuits

KVL to the circuit of Fig. 4.14 and taking the clockwise direction as being
positive,

V- VL- VR = 0 (phasorially)

so that

V = VL + VR (phasorially) = IXL + IR (phasorially)

Figure 4.15

It-

VR

I (reference)

This phasor addition is shown in Fig. 4.15. The current is taken to be the
reference phasor because it is common to both elements. From Equation (4.12),
the voltage drop (IR) in the resistance R is in phase with the current (I). From
Equation (4.13), the voltage drop (IXL) in the inductive reactance XL is 90 ~
ahead of the current. These two voltage drops are then summed to give the
source voltage V.
From the phasor diagram we see that, by Pythagoras' theorem,
V 2 --- VR 2 -+- VL 2 SO that

V2= (/R) 2 + (IXL) 2= I2(R2 + XL 2)

Taking the square root of both sides we get

V = I~/(R 2 + XL 2) = IZ (4.17)

where Z = ~v/(R2 + XL 2) and is called the impedance of the circuit. Since
Z = VII its unit is the volt per ampere (the ohm).
The phasor diagram is now as shown in Fig. 4.16(a) and if we divide all three
phasors by I we obtain the diagram of Fig. 4.16(b) which is called an impedance
triangle. The angle ~b is the phase angle of the circuit, and from the geometry of
the triangle we see that R/Z = cos ~b, XL/Z = sin ~b and XL/R = tan ~b.

~IXL

9 r

IR

I (reference)

Figure 4.16

r j

R

(a) (b)

XL

Example 4.7

In the circuit of Fig. 4.14, R = 5 11, l = 50 mH and v = 100 sin 628t. Determine
(1) the inductive reactance of the circuit, (2) the impedance of the circuit, (3)
the current drawn from the supply, (4) the phase angle of the circuit.