, which includes the derived resistances. Use formula (A.4): RT = 480 + 1277.14 + 600 + 900(S2) = 3257.14 S2. To calculate the current flowing out of the battery, use Ohm's law [Eq. (A.2)]: I = E/R = 48/3257.14 = 0.0147 A or 14.7 mA. A.5 ELECTRIC POWER IN dc CIRCUITS Batteries store chemical energy. When the battery terminals are connected to supply emf to an electric circuit, the battery chemical energy is converted to electrical energy. This A.6 INTRODUCTION TO ALTERNATING CURRENT CIRCUITS manifests itself as power (work per unit time). The unit of power is the watt (W). When we pay our electric bill, we pay for kilowatt-hours of expended electric power. The resistors in Figures A.2-A.8 dissipate power in the form of heat. In fact, not only are resistors rated for their resistance in ohms, but also for their capability to dissipate heat measured in watts. Let us now examine the electric power dissipated by a resistor or other ohmic device.4 Let power, expressed in watts, be denoted by the notation P. Then: P = EI. (A.14) Stated in words: In an electric circuit, if we multiply the electromotive force in volts by the current in amperes, we have an expression for the power in watts [Eq.(A.14)]. The watt may, therefore, be defined as the power expended in a circuit having an electromotive force of one volt and a current of one ampere. There are two variants of Eq. (A.14) by simple substitution of variants of Ohm's law [Eqs. (A.1), (A.2), and (A.3)]: (A.15) P = EI = IR(I) = 12R. (A.16) Example 1. The current flowing through a 1000-S resistor is 50 mA. What is the power dissipated by the resistor? Convert mA to A or 0.05 A. Use Eq. (A.16): = (0.05)2 X 1000 =2.5 W. Example 2. The potential drop across a 600-S2 resistor is 3 V. What power is being dissipated by the resistor? Use Eq. (A.15): P = W or 15 mW. A.6 INTRODUCTION TO ALTERNATING CURRENT CIRCUITS A.6.1 Magnetism and Magnetic Fields The phenomenon of magnetism was known by the ancient Greeks when they discovered a type of iron ore called magnetite. Moorish navig