G= 



On Measuring Resistances with Wheatstone's Diagram. 365 



Kirchhoff, the following six equations, which are independent of 

 one another: — 



A=B + G, 

 C = D + G, 

 F = A + D, 

 aA+gG-dT) = ) 

 fF + ak+gG + cC=E, 

 and gG + cC-bB=0. 



By eliminating A, B, C, D, and F we obtain a single equation 

 containing the electromotive force E, the current G in the gal- 

 vanometer, and the six resistances a, b, c, d, g } and/. If we now 

 develope G from this equation, we have 



E 



(c + d)(a + b)+f(a + b + c + d) Jb + c)(a + d)+ab(c + d)+cd{a + b )' 

 bd-ac +/ bd—ac 



by putting bd—ac = 0, we have G=0, which is the known law 

 of Wheatstone's diagram. 

 Let us substitute 



{c + d)(a + b)+f{a + b + c + d)=Y, 



bd—ac = u, 



f{b + c)(a + d) + ab(c + d)+cd(a + b)=W, 



and multiply the equation by U, the number of turns in the gal- 

 vanometer-coil, we obtain its magnetic moment, which may be 

 called Y, 



Y==Ea -== — ™r« 



The space to be filled with wire of known and constant conduc- 

 tivity being given, and calling q the sectional area of this wire, we 

 have between U, q, and g the following two equations : — 



TT— const * 



Q ' 



U const. 

 9 = 



or therefore 



Q 

 U= const. ts /g 9 



and consequently 



Y=E«._^_. const. =E« ^9 



gV + W #V + W 



