REAL GENERATORS 



Example of application — Find the current in the 2 ohms resistance of the 

 bridge circuit in Figure 2.15 if a constant- vohage generator of e.m.f. 10 volts is 

 connected across A-B. 



A/VSA- 



tigiire 2.18 



Figure 2.17 



First, remove the 2 ohms resistance {Figure 2.18). We now have two 

 potential dividers and the bridge can be solved by the rules for series resis- 

 tances: 



The resistance in the path A-C-B is 7 ohms ; 



therefore the current in the path A-C-B is 10/7 amps; 



therefore the potential difference A-C is 3 X (10/7) volts. 

 By a similar process the P.D. A-D is 5 X (10/1 1) volts. So the P.D. C-D is 

 {5 X (10/11)} — {3 X (10/7)} and this is our E' . To find /•', bearing in mind 



3Q 



l-^/^A/v^ 





sn 



B 



r' 



-VWVVV^ 



'0I« 



(f'l)n 



JIO 



11 



)-(3 



JIO 

 7 



volts 



/?>2Q 



6Q 



Figure 2.19 Figure 2.20 



that the generator has no internal resistance, we re-draw the bridge as in 

 Figure 2.19. 'Looking in' at C-D, we see a resistance of 3 ohms in parallel 

 with 4 ohms, plus 5 ohms in parallel with 6 ohms. 



3X4 , 5x6 



r = 



12 30 

 7 "^ 11 



3+4 ' 5+6 



The equivalent circuit is therefore Figure 2.20 and the current through R 

 may be quickly found. 



G L 



Figure 2.22 



Figure 2.23 



Figure 2.21 



Potentiometer control of power 



In Figure 2.21 we have a generator G feeding some kind of load L. We 

 often wish to control the amount of power being supplied to L. If G is of 

 the constant-voltage type, such as the electric supply mains, and L is, say, a 

 lamp, an appropriate method would be a variable series resistance {Figure 

 2.22) or 'rheostat' to 'drop' some of the voltage. Similarly, if G is of the 



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