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CHAPTER 2 Electric Current. Ohm s Law
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then the algebraic form of Ohm s law becomes I kV R 10
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in which k is a constant of proportionality. Equation (10) says that current I is directly proportional to potential di erence V and inversely proportional to resistance R. That is, the greater V the greater is I, and the greater R the less is I. The unit of resistance is called the ohm; we de ne that a conductor has 1 ohm of resistance if 1 ampere of current ows when a potential di erence of 1 volt is applied to the conductor. Thus, if V 1 and R 1, then by de nition I 1, and eq. (10) becomes, 1 k 1 = 1 , which can be true only if k 1. Therefore, if we express V in volts, I in amperes, and R in ohms, then eq. (10) becomes the basic OHM S LAW I V R 11
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which states that amperes is equal to volts divided by ohms. It follows that Ohm s law can also be written in either of the forms R V=I and V RI 13 12
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Equation (12) says that ohms is equal to volts divided by amperes, while eq. (13) makes the equivalent statement that volts equals ohms times amperes. Equations (11), (12), and (13) are basic to electrical and electronic engineering, and should be committed to memory. Now let s consider the POWER developed by the battery in Fig. 22. To do this, let us begin with a brief review of some basic concepts, as follows. First, we have the idea of energy, which is measured in terms of capacity to do work, which is measured in joules. In mechanics, when a force of F newtons acts through a distance of L meters, the agency supplying the force does an amount of work, W, equal to FL joules, that is, W FL. It is important, now, to notice that there is no time requirement in the de nition W FL. Thus, suppose in a certain case that FL joules of work must be done. Such a simple requirement is satis ed regardless of whether the work is done in 1 minute or in 10 minutes. Actually, however, in plain language we know that a more powerful source of energy is required to do the work in 1 minute than in 10 minutes. For example, a small boy might, with the aid of a system of pulleys, raise a 100-pound weight 1 foot o the oor in, say, 60 seconds. An adult, however, might, without having to use pulleys, be able to do the same thing in, say, 6 seconds. The same amount of work (100 foot-pounds) is done in both cases, but the adult, while working, is delivering energy to the system 10 times as fast as the boy is capable of doing. Thus, the time rate of doing work is important in practical engineering. In the mks system time rate of doing work is expressed in joules per second, which is given the special name watts, in honor of the early engineer James Watt. Thus we have the de nition time rate of doing work joules per second watts 14
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CHAPTER 2 Electric Current. Ohm s Law
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Now let P denote the power developed by the battery in Fig. 22. We wish to show that the power, P watts, is equal to the battery voltage times the current I; that is, we wish to show that P VI. To do this, we make use of the basic de nitions, volts joules per coulomb and current coulombs per second, and then manipulate the units as if they were ordinary algebraic quantities, thus VI joules coulombs joules joules per second watts; thus; coulombs seconds seconds P VI 15
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which says that the power in watts produced by a battery of V volts when delivering a current of I amperes is equal to VI; that is, power in watts is equal to volts times amperes. In Fig. 22, the power VI, produced by the battery, is delivered, by means of connecting copper wires, to the load resistance R. If R is, for example, the lament of a light bulb, then the power supplied to R will be converted into heat energy and into radiant energy in the form of visible light. Or, if R is an electric motor, the majority of the battery output will, hopefully, be converted into useful mechanical energy, with the relatively small balance being lost in the form of heat energy. The power output of the battery, given by eq. (15), is of course the same as the power delivered to and consumed by the load resistance R. We can therefore use eqs. (11) and (13) to write equations for the power delivered to a resistance of R ohms, as follows. First, using eq. (11), eq. (15) becomes P V V=R , so that P V2 R 16
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