ABSTRACT
In an a.c. circuit containing pure capacitance only (see Figure 27.3(a)), the current IC leads the applied voltage VC by 90◦ as shown in the phasor diagram of Figure 27.3(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 27.3(c). The impedance Z of the circuit is given by
Z = VC∠−90 ◦
IC∠0◦ = VC
IC ∠−90◦ = XC∠−90◦ or −jXC
where XC is the capacitive reactance given by
XC = 1 ωC
= 1 2πfC
ohms
where C is the capacitance in farads.[ Note:−jXC = −j
ωC = −j (j)
ωC(j)
= −j 2
jωC = −(−1)
jωC = 1
jωC
]
In an a.c. circuit containing resistance R and inductance L in series (see Figure 27.4(a)), the applied voltage V is the phasor sum of VR and VL as shown in the phasor diagram of Figure 27.4(b). The current I lags the applied voltage V by an angle lying between 0 ◦ and 90◦ – the actual value depending on the values of VR and VL, which depend on the values of R and L. The circuit phase angle, i.e. the angle between the current and the applied voltage, is shown as angle φ in the phasor diagram. In any common to all components erence phasor in Figure may be superimposed on in Figure 27.4(c), where it form the supply voltage V
V = V
Figure 27.5(a) shows the from the phasor diagram angle Oab). If each side divided by current I then Figure 27.5(b) is derived. be superimposed on the
Figure 27.5(c), where it may be seen that in complex form the impedance Z is given by:
Z= R+ jXL
(3+j4) means that the resistance is 3 and the inductive reactance is 4. In polar form, Z=|Z|∠φ where, from the impedance triangle, the modulus of impedance |Z|= √(R2+X2L) and the circuit phase angle φ= tan−1(XL/R) lagging.
