Principles of naval engineering - Historic Naval Ships Association

PRINCIPLES OF NAVAL ENGINEERINGSince there is only one path for current inthe series circuit, the total current is the samein all parts of the circuit. Dividing both sides ofthe voltage equation by the common factor I, anexpression is derived for the total resistance ofthe circuit in terms of the resistances of theindividual devices:RlAA/VW-5nRt = Rl- R2 + R3Therefore, in series circuits the total resistanceis the sum of the resistances of theindividual parts of the circuit.In the example of figure 20-6, the totalresistance is 5+10+15= 30 ohms. The totalcurrent may be found by applying the equation^t 30 ^TI). = on = 1 ampereRt"^"The power absorbed by resistor R^ is I^R^,or 1^x5=5 watts. Similarly, the power absorbedby R2 is 1^ X 10 = 10 watts, and the power absorbedby R3 is l2 X 15 = 15 watts. The totalpower absorbed is the arithmetic sum of thepower of each resistor, or 5 + 10 + 15 = 30 watts.The value is also calculated by Pt = Etl^ =30 X 1 = 30 watts.Parallel CircuitsThe parallel circuit differs from the simpleseries circuit in that two or more resistors, orloads, are connected directly to the same sourceof voltage. There is accordingly more than onepath that the electrons can take. The more paths(or resistors) that are added in parallel, the lessopposition there is to the flow of electrons fromthe source. This condition is opposite to the effectthat is produced in the series circuit whereadded resistors increase the opposition to theelectron flow.As may be seen from figure 20-7, the samevoltage is applied across each of the parallelresistors. In this case the voltage applied acrossthe resistors is the same as the source voltage,Figure 20-7.— Parallel electric circuit.13.16resistance of the branch, the higher will bethe current through that branch. The individualcurrents can be foundby the application of Ohm'slaw to the individual resistors. Thus,andand^sT 30 „I1 = — = 3— = 6 amperes^1s 30 „I9 = — = rr- = 3 amperes10RrE3 Rq30301 ampereThe total current. It, of the parallel circuitis equal to the sum of the currents through theindividual branches. This, in slightly differentwords, is Kirchhoff's law. In this case, the totalcurrent isT = Ij + I2 + I„= 6 + 3 + 1 = 10 amperesCurrent flows from the negative terminal ofthe source to point a where it divides and passesthrough the three resistors to point band back tothe positive terminal of the voltage source. Theamount of current flowing through each individualbranch depends on the source voltage and on theresistance of that branch— that is, the lower theIn order to find the equivalent, or total,resistance (R^) of the combination shown infigure 20-7, Ohm's law is used to find each ofthe currents (I^, Ii, I2, and Ig) in the precedingformula. The total current is equal to the sumof the branch currents. Thus,498