Page:Elementary Text-book of Physics (Anthony, 1897).djvu/375

§ 305] as negative which are opposite to the direction of motion. Further, those electromotive forces are positive which tend to set up positive currents in their respective branches. With these conventions Kirchhoff's laws may be stated as follows:

1. The algebraic sum of all the currents meeting at any point of junction of two or more branches is equal to zero. This iirst law is evident, because, after the current has become steady, there is no accumulation of electricity at the junctions.

2. The sum, taken around any number of branches forming a closed circuit, of the products of the currents in those branches and their respective resistances is equal to the sum of the electromotive forces in those branches. This law can easily be seen to be only a modified statement of Ohm's law.

These laws may be illustrated by their application in a form of apparatus known as Wheatstone's bridge. The circuit of the Wheatstone's bridge is made up of six branches. An end of any branch meets two, and only two, ends of other branches, as shown in Fig. 87. In the branch 6 is a voltaic cell with an electromotive force $$E.$$ In the branch 5 is a galvanometer which will indicate the presence of a current in that branch. In the other branches are conductors, the resistances of which may be called respectively $$r_{1}, r_{2}, r_{3}, r_{4}.$$

From Kirchhoff's first law the sum of the currents meeting at the point $$C$$ is $$i_{1} + i_{2} + i_{5} = 0,$$ and of those meeting at the point D is $$i_{3} + i_{4} + i_{5} = 0.$$ By the second law, the sum of the products $$ir$$ in the circuit $$ADC$$ is $$i_{1}r_{1} + i_{3}r_{3} + i_{5}r_{5} = 0,$$ and in the circuit $$DBC$$ is $$i_{4}r_{4} + i_{2}r_{2} + i_{5}r_{5} = 0,$$ since there are no electromotive forces in those circuits. If we so arrange the resistances of the branches 1, 3, 3, 4 that the galvanometer shows no deflection, the current $$i_{5}$$ is zero, and these equations give the relations $$i_{2} = -i_{4}, \, i_{1}r_{1} = -i_{3}r_{3}, \, i_{2}r_{2} = -i_{4}r_{4}.$$ From these four equations