350-352] Currents in a Network 



let the current through ABD be C. Then, by Ohm's Law, 



305 



so that 



R 2 v-V B * 



From a similar consideration of the flow in ACD, we obtain 



R 3 V A -v 



so that we must have 



R 



.(272), 



as the condition to be satisfied between the resistances when there is no 

 current in BC. 



Clearly by adjusting the bridge in this way we can determine an unknown 

 resistance R l in terms of known resistances R^, R 3) R 4 . In the simplest 

 form of Wheatstone's bridge, the line ACD is a single uniform wire, and the 

 position of the point can be varied by moving a " sliding contact " along 

 the wire. The ratio of the resistances R 3 : R 4 is in this case simply the ratio 

 of the two lengths AC, CD of the wire, so that the ratio ^ : R 2 can be found 

 by sliding the contact C along the wire ACD until there is observed to be 

 no current in BC, and then reading the lengths AC and CD. 



EXAMPLES OF CURRENTS IN A NETWORK. 



I. Wheatstone's Bridge not in adjustment. 



352. The condition that there shall be no current in the 

 in fig. 98 has been seen to be that given by equation (272). 



B 



bridge " BC 



Suppose that this condition is not satisfied, and let us examine the flow 

 of currents which then takes place in the network of conductors. Let the 

 conductors AB, BD, AC, CD as before be of resistances R 1 , R 2 , R 3 , R 4 , and 

 let the currents flowing through them be denoted by x lt x 2 , x 3 , % 4 . Let the 

 bridge BC be of resistance R b , and let the current flowing through it from 

 B to C be x b . 



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