of the two resistors in parallel. Applying Ohm's law, we obtain R = E/I. Substitute Eq. (A.5). We then have R = =E/(I1+12). (A.6) However, I = E/R1 and I2 = E/R2. As a consequence, we have R = E/[(E/R1)+(E/R2)] (A.7) Divide Eq. (A.7) through by E and we now have R = =1/[(1/R1)+(1/R2)]. (A.8) Simplify the compound fraction and we have R = R1 x R2/(R1+ R2). (A.9) Example. There are two resistors connected in parallel. Their values are 500 So and 700 S2, respectively. What is the value of the equivalent combined resistance? Use Eq. (A.9). R = 500 x 700(500 + 700) = 350,000/(500+700) = 350,000/1200 = = = 291.6 One self-check for resistances in parallel is that the equivalent combined resistance must be smaller than the value of either resistor. Figure A.6 shows a group of three resistors in parallel with an emf source that is a battery. We encourage the use of a short-cut when there are more than two resistors in parallel. We introduce a new term, conductance. Conductance, G, is the inverse of resistance; stated in an equation: R = 1/G. (A.10) The unit of conductance is the mho, which the reader will note is "ohm" spelled backwards. To solve the equivalent total resistance problem of Figure A.6, convert each resistance into its equivalent conductance, sum the values, and invert the sum. Example. The values of the resistances in Figure A.6 are 2000, 2500, and 3000 S2; the closed circuit emf of the battery is 24 V. What is the value of the current flowing out of the battery? Use Eq. (A.10) and convert each resistance value to its equivalent conductance REVIEW OF FUNDAMENTALS OF ELECTRICITY WITH TELECOMMUNICATION APPLICATIONS Figure A.6 Model of a circuit with three resistors in parallel. and sum: G=G + G2+G3 = 1/2000 + 1/2500 1/3000 = 0.0005 + 0.0004 + 0.00033 = 0.00123 mhos. (A.11) Use the inverse of Eq. (A.11) or R = 1/G; then R = 1/0.00123=813 S. We know the voltage source is 24 V and the circuit resistance is 813 S2. We now can apply Ohm's law to calculate the current flowing out of the battery. Use Eq. (A.2) or I 