RESISTANCES 



For example, the effective resistance of the network in Figure 2.7 is easily 

 found, as follows. 



We first combine the resistances from AtoB into one of 3 + 1'5 ohms = 

 4-5 ohms. Similarly, between C and D we have 6 + 3 = 9 ohms. The net- 

 work simplifies to Figure 2.8. We now consider the 4-5 ohms and 9 ohms 



Figure 2.7 



Figure 2.8 



5n 



fl,D 



Figure 2.9 



resistances in parallel; they are equivalent to a single resistance of (9 X 

 4-5)/(9 + 4-5) = 3 ohms {Figure 2.9). Next we combine the newly found 

 3 ohms with the 7 ohms yielding Figure 2.10. Finally we are left merely with 

 5 ohms in parallel with 10 ohms,_which reduces to a single equivalent resis- 

 tance of (5 X 10)/(5 + 10) = 3-33 ohms (Figure 2.11). 



To solve all the voltages and currents which flow in Figure 2.7 when a 

 generator is connected across the terminals, one quick way is to assume part 



FE 



3-33ft 



G B,D 



Figure 2.10 



G,B,D. 

 Figure 2.11 



of the answer; e.g. that the current in AB is one amp downwards. Since 

 the currents divide inversely as the ratio of parallel resistances, the current 

 CD must be | amp downwards. Thus the current in E-AC is l^amps 

 downwards. In Figure 2.10, if we have 1| amps flowing from E to BD, the 

 current down F-G must be 3 amps, and the total current suppUed to the 

 network must be 4| amps. If we are told that the generator actually supplies 

 13 amps to the network, then the actual currents flowing in the various 

 parts of the network are found merely by multiplying the assumed values 

 by 13/4|. Finally the potential differences across the resistances are worked 

 out by multiplying the current through each resistance by its value in ohms. 



10 



