Xc (A.34a) =2xfL-1x10°/2xfC (A.34b) A word about signs: If the calculated value of X is positive, inductive reactance predom- inates, and if negative, capacitive reactance predominates. A.8 ALTERNATING CURRENTS 2.16 uF 1020 Hz Figure A.22 A simple ac circuit displaying capacitance only. Example. Figure A.23 illustrates an example of a circuit with capacitance and inductance in series. The inductance value is 400 mH and the capacitance is 500 nF. The source emf is 20 V. The frequency is 1020 Hz. Calculate the current in the circuit. Assume the resistance is negligible. Calculate the combined reactance X of the circuit using formula (A.34). Convert units for capacitance to microfarads, and convert the units of inductance to henrys. That is 0.5 uF and 0.4 H. X 3.14159 X 1020 X 0.4 - 1 106/2 X 3.14159 X 1020 X 0.5 = 2563.537 - 1,000,000/3204.42 = 2563.537 - 312.068 = 2251.469 S2. Apply Ohm's law variant to calculate current: I=E/X = 20/2251.469 = 0.00888 A or 8.88 mA. 500 nF 0.4 H 1020 Hz Figure A.23 A simple ac circuit with a coil (inductance) and capacitor in series. REVIEW OF FUNDAMENTALS OF ELECTRICITY WITH TELECOMMUNICATION APPLICATIONS A.8.3 Calculating Impedance When we calculate impedance (Z), we must take into account resistance. All circuits are resistive, even though in some cases there is only a minuscule amount of resistance. We first examine the two reactive possibilities; that is, a circuit with inductive reactance and then a circuit with capacitive reactance. For the case with inductive reactance, we have (A.35a) Substituting, we obtain (A.35b) For the case with capacitive reactance, we have (A.36a) Substituting, we obtain (A.36b) We can also state the impedance: Substituting, we obtain (A.37) Example. Figure A.24 illustrates a simple ac circuit consisting of resistance (100 S2), capacitance (2.16 uF), and inductance (400 mH) in series. The frequency is 1020 Hz and 2.16 uF 100 0 400 mH 1020 Hz Figure A.24 A simple ac circuit with resistance, capacitance, and inductance in series, A.10 RESONANCE the ac