attery emf). By direct inference from the figure we can see the equivalent capacitance of the three capacitors is the sum of each of the individual capacitances or Cog = C1+C2+C3. (A.23) Suppose the values of the capacitors were 2, 3, and 4 nF, respectively. What would the equivalent capacitance be? From Eq. (A.23), the value would be 9 nF. Capacitors in series (see Fig. A.14) are handled in a similar manner as resistances in parallel. The total equivalent capacitance is 1/Ceq=1/C1+1/C2+1/C3. (A.24) Suppose, again, the values of the three capacitors were 2, 3, and 4 nF. What would the equivalent capacitance be? 1/Ceq=1/2+1/3+1/4 = 6/12+4/12+3/12= = 13/12, Ceq = 12/13 = 0.923 nF. Figure A.14 Capacitors in series. REVIEW OF FUNDAMENTALS OF ELECTRICITY WITH TELECOMMUNICATION APPLICATIONS A.7.2 RC Circuits and the Time Constant An RC circuit is illustrated in Figure A.15. The capacitor in the figure remains uncharged until the switch is closed. To charge it we throw the switch S to a closed position. The circuit consists of a resistor (R) and capacitor (C) in series. When we close the switch, there is a surge of current whose intensity decreases with time as the charge of the capacitor builds up. Figure A.16 shows the change in the value of the current with time after the switch closed. There is a corresponding change in the voltage drop across the capacitor C. This is illustrated with curve E. The instantaneous value of current, I, is determined solely by the value of resistor, R. The total voltage drop is then across R and the drop across C is zero. As the capacitor begins to charge, however, the voltage drop across C gradually builds up, the current decreases, and the voltage drop across R decreases correspondingly. When the capacitor reaches full charge and the current has fallen to a negligible value, the total voltage drop is now across the capacitor. Remember that at all times the sum of the voltage drops across R and C must be equal and opposite to the impressed voltage. Where does time come into the picture 