164 Drs. Crehore and Squier on the Currents 



I 1)3 equation (1). The line OD then represents the current 

 flowing in the circaits (1) and (3). A similar calculation 

 may be made for the branches (2) and (4), and another current 

 OD / drawn upon the diagram. Having determined the two 

 branch carrents OD and OD', the current in the main line is 

 the geometrical sum of these currents, represented by the 

 line OR. 



The potential-difference between the points C and D of the 

 bridge is by hypothesis zero, and that between the terminals 

 A and B is E. It remains to find the potential-differences 

 between the points A and C, A and D, C and B, and D and B. 

 Draw a perpendicular from A upon the line OD, then OC 

 represents the whole of that component of the total E.M.F., 

 E, which goes to overcome the resistance of the two circuits 

 (1) and (3). This E.M.F. following Ohm's law is equal to 

 the product of the current and resistance. Hence 

 OC = (R l + B 3 )I 13 . Similarly a point C may be found upon 

 the line OD', and OC'= (R 2 + R 4 ) I 2>4 . These lines OC and 

 OC may then be divided into two parts proportional to the 

 resistances R 1? R 3 and R 2 , R 4 . Such a division gives the 

 points H on OC, and K on OC, so that OH = Ri I lt8 ; HC = 

 R 3 I L3 ; OK = R 2 I 2j4 ; and KC = R 4 I 2 , 4 . 



Just as OC represents the component of the total E.M.F. , 

 E, which is in the direction of the current, and may be called 

 the " power E.M.F./' so CA represents that component which 

 is at right angles to the current, and is called the " reactive 

 E.M.F." This part of the total E.M.F. is due to the presence 

 of the magnetic field in the circuits, and it may be divided 

 up in a similar manner into parts which represent the sepa- 

 rate effects of the coils (1), (2), (3), and (4). The line CA 

 is equal to (Si, 3 La)) I li3 , and may be divided at M into the two 

 parts CM^Li © I 1>3 and MA = L 3 g>I 1)3 . Similarly CA is 

 divided at N into two parts CN = L 2 &) I 2)4 and NA= L 4 co I 2A . 



To find the potential-difference at the terminals of any coil 

 as AC (fig. 1) we need only combine the component E.M.F.' s 

 for that coil, one in the direction of the current and the other 

 at right angles to it. This gives for circuit AC (1) the line 

 OH = Ri I 1<3 in the direction of the current, and HP=--L 1 a> I 1>3 

 at right angles to the current to be geometrically added 

 together, making the resultant OP as the potential-difference 

 at the terminals of the coil AC (1). A similar process gives 

 OQ for circuit AD (2), and PA for CB (3), and QA for 

 DB(4). 



Fig. 2 is drawn to represent the case of a divided circuit, 

 which becomes the Wheatstone's bridge by joining a circuit 

 across between the branches. 



If such a connexion were made, the arrangement of the 



