ABSTRACT
XL = 2π fL = 2π(50)(40 × 10−3) = 12.57
Current, I = V XL
= 240 12.57 = 19.09 A
(b) Inductive reactance,
XL = 2π(1000)(40 × 10−3) = 251.3
Current, I = V XL
= 100 251.3
= 0.398 A
In a purely capacitive a.c. circuit, the current IC leads the applied voltage VC by 90◦ (i.e. π/2 rads). See Fig. 15.4
In a purely capacitive circuit the opposition to the flow of alternating current is called the capacitive reactance, XC
XC = VC IC
= 1 2πfC
where C is the capacitance in farads. XC varies with frequency f as shown in Fig. 15.5
Problem 3. Determine the capacitive reactance of a capacitor of 10µF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz
(a) Capacitive reactance XC = 12π fC
= 1 2π(50)(10 × 10−6)
= 10 6
2π(50)(10) = 318.3
(b) XC = 12π fC = 1
2π(20 × 103)(10 × 10−6)
= 10 6
2π(20 × 103)(10) = 0.796
Hence as the frequency is increased from 50 Hz to 20 kHz, XC decreases from 318.3 to 0.796 (see Fig. 15.5)
Problem 4. A capacitor has a reactance of 40 when operated on a 50 Hz supply. Determine the value of its capacitance.
