Introduction to Electric Circuits

4.3 Series a.c. circuits 79

Solution

1 From Equation (4.15) the inductive reactance

X L = toL = 628 • 50 x 10 -3 = 31.4 1)

2 From Equation (4.17) the impedance

Z = X/(R 2 + XL 2) = X/(52 + 31.42) = 31.79 1~

3 The current I = V/Z = 100/31.79= 3.15 A.

4 The phase angle 6 = tan-1 (XL/R) =tan-' (31.4/5) - 80.95 ~

Series R C circuits

vt(

I R

VR

)

C

-41

Vc

Figure 4.17

Applying KVL to the circuit of Fig. 4.17 and taking the clockwise direction to

be positive we have

V- Vc- VR--0 (phasorially)

so that

V- V R .n t.- V C -IR + IXc (phasorially)

With I as the reference phasor, the phasor diagram is shown in Fig. 4.18(a). In

VR = IR I (reference) R

Vc = IXc

V=IZ

Xc

(a) (b)

Figure 4.18

this diagram VR (-IR) is in phase with the current (in accordance with
Equation (4.12)), and Vc (=IXc) lags the current by 90 ~ (in accordance with
Equation (4.14)). Now

V 2- (/R) 2 + (IXc) 2= 12(R2 + Xc 2)

so that

v- W(,e ~ + x~ ~) = IZ

where Z - ~v/(R 2 + Xc 2) and is called the impedance of the circuit.