nce and capacitive reactance. Inductive Reactance. As we learned earlier, the value of current in an inductive circuit not only varies with the inductance but also with the rate of change of current magnitude. This, of course, is frequency (f). Now we can write an expression for a circuit's inductive reactance, which we will call XL. It is measured in ohms. XL 2rfL, (A.32) where L is the circuit's inductance in henrys. Example. Figure A.21 shows a simple inductive circuit where the frequency of the emf source is 1020 Hz at 20 V, the inductance is 3.2 H. Calculate the effective current through the inductance. There is negligible resistance in the circuit. Use Eq. (A.32). 2 x 3.14159 X 1020 X 3.2 = 20,508.3 S2; I = 20/20, 508.3 = 0.000975 A or 0.975 mA. In this circuit the voltage lags the current by 90°. REVIEW OF FUNDAMENTALS OF ELECTRICITY WITH TELECOMMUNICATION APPLICATIONS 3.2 H 1020 Hz Figure A.21 An ac circuit with inductance only. Capacitive Reactance. Capacitive reactive has the opposite behavior of inductive reac- tance. In this case, the current lags the voltage by 90°. Also, as the frequency increases, the capacitive reactance decreases, whereas with inductive reactance, as the frequency increases, the reactance increases. The following is an expression to calculate capaci- tive reactance: Xc=-1/2rfC (A.33a) where f is in hertz and C is in farads. The more customary capacitance unit is the microfarad (uF). When we use this unit of capacitance, the formula in Eq. (A.33a) becomes: (A.33b) Example. Figure A.22 illustrates a capacitive reactance circuit with a standard capacitor of 2.16 uF and an emf of 20 V at 1020 Hz. Calculate the current in amperes flowing in the circuit. Use formula (A.33b): Xc = x 106/2 x 3.14159 x 1020 x 2.16 = -72.24 S2, = E/Xc I = -20/72.24 = -0.277 A (minus sign means leading current). Circuits with Combined Inductive and Capacitive Reactance. To calculate the combined or total reactance when an inductance and a capacitance are in series, the following formula is applicable: (=XL+