PRINCIPLES OF NAVAL ENGINEERING 



Since there is only one path for current in 

 the series circuit, the total current is the same 

 in all parts of the circuit. Dividing both sides of 

 the voltage equation by the common factor I, an 

 expression is derived for the total resistance of 

 the circuit in terms of the resistances of the 

 individual devices: 



Rt = Rl - R2 + R3 



Therefore, in series circuits the total re- 

 sistance is the sum of the resistances of the 

 individual parts of the circuit. 



In the example of figure 20-6, the total 

 resistance is 5+10+15= 30 ohms. The total 

 current may be found by applying the equation 



T ^t 30 ^ 



I). = on = 1 ampere 



Rt "^" 



Rl 



■AA/VW- 

 5n 



13.16 



Figure 20-7.— Parallel electric circuit. 



The power absorbed by resistor R^ is I^R^, 

 or 1^x5=5 watts. Similarly, the power absorbed 

 by R2 is 1^ X 10 = 10 watts, and the power ab- 

 sorbed by R3 is l2 X 15 = 15 watts. The total 

 power absorbed is the arithmetic sum of the 

 power of each resistor, or 5 + 10 + 15 = 30 watts. 

 The value is also calculated by Pt = Etl^ = 

 30 X 1 = 30 watts. 



Parallel Circuits 



The parallel circuit differs from the simple 

 series circuit in that two or more resistors, or 

 loads, are connected directly to the same source 

 of voltage. There is accordingly more than one 

 path that the electrons can take. The more paths 

 (or resistors) that are added in parallel, the less 

 opposition there is to the flow of electrons from 

 the source. This condition is opposite to the ef- 

 fect that is produced in the series circuit where 

 added resistors increase the opposition to the 

 electron flow. 



As may be seen from figure 20-7, the same 

 voltage is applied across each of the parallel 

 resistors. In this case the voltage applied across 

 the resistors is the same as the source voltage, 



resistance of the branch, the higher will be 

 the current through that branch. The individual 

 currents can be foundby the application of Ohm's 

 law to the individual resistors. Thus, 



T ^s 30 „ 



I1 = — = 3— = 6 amperes 



^1 



and 



and 



s 30 „ 

 I9 = — = rr- = 3 amperes 



Rr 



E 

 3 Rq 



10 



30 

 30 



1 ampere 



The total current. It, of the parallel circuit 

 is equal to the sum of the currents through the 

 individual branches. This, in slightly different 

 words, is Kirchhoff's law. In this case, the total 

 current is 



T = Ij + I2 + I„ = 6 + 3 + 1 = 10 amperes 



Current flows from the negative terminal of 

 the source to point a where it divides and passes 

 through the three resistors to point band back to 

 the positive terminal of the voltage source. The 

 amount of current flowing through each individual 

 branch depends on the source voltage and on the 

 resistance of that branch— that is, the lower the 



In order to find the equivalent, or total, 

 resistance (R^) of the combination shown in 

 figure 20-7, Ohm's law is used to find each of 

 the currents (I^, Ii, I2, and Ig) in the preceding 

 formula. The total current is equal to the sum 

 of the branch currents. Thus, 



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