136 BKIDGMAN. 



current density is to be high. In this branch we suppose that 

 Ohm's law does not hold, but the relation between current and E.M.F. 

 is given by a curve of the form shown in Figure 2. The other arms of 

 the bridge, Ro, R3, and Ri are made of larger wire, in which the current 

 density is always small, and hence their resistance is ohmic. Assume 

 for the moment that it is possible to balance the bridge simultaneously 

 for D.C. and A.C. Now let a heavy direct current h flow through z 

 and Ro and a direct current I3 through R3 and Ri. Also let a small 

 alternating current h sin w/ flow through x and R2 and is sin w< flow 

 through R3 and Ri. The difference of potential between the ends of 

 X is 



Ii tan 6 + ?'i tan 6' sin wt 



The potential difference across the galvanometer is 



(/i tan 6 + i\ tan 6' sin ut) — {I3 + is sin oi)t) R3. 



Since the extremities of Ro and Ri join, their potentials are equal, or 



/i (tan + i?2) + h (tan 6' + R.2) sin o^t = {1 3 + (3 sin w^ {R3 + ^4). 



This splits into two equations 



7i tan 6+ hR2= I3 (i?3 + Ri) (1) 



ii tan d' + h /?2 = is (Rs + Ri) (2) 



Now if the galvanometer is balanced for D.C. the constant part of 

 the potential difference across its terminals must vanish, or 



7i tan - /a i?3 = 0. 



Combined with (1) above we get 



h R2 - hRi= 0, 

 or, dividing 



tan d R3 



Ro Ri 



(3) 



In the same way, if the bridge is balanced for A.C. the alternating 

 part of the potential difference across the galvanometer vanishes, or 



(4) 



