_=326. Discussion; Equipotential Points.=_ Since one end of the G-s W has a higher, and its other end has a lower potential than M P, there must be, somewhere on it, a point at which the potential is the same as at M P. This place is quickly found by sliding the free end of wire, 3, along, pressing K occasionally, until A G shows that no current tends to pass through it in either direction, when the current passes from C to Z through the two branches of the divided circuit. This point and M P are called _equipotential points_. If the resistance of the part, X, be increased, it should be evident that the part of the bridge-wire, B, should be also increased to find a point having the same potential as M P; that is, the end of 3 should be moved towards C. We have, in the bridge-wire, a simple means of varying the resistance of its parts, A and B. =327. Use of Wheatstone's Bridge.= It will be found, upon trial, if we put a resistance of 2 ohms in place of R, Fig. 102, and 2 ohms in place of X, that the free end of wire 3 will have to be at the center of the bridge-wire in order to get a "balance"; that is, to find the place where A G is not affected. No matter what the resistance of R and X are, provided they are equal, this will be true. The value of both A and B, on the scale, will be 5 whole spaces, no tenths. From this we see that A: B:: R: X, which reads A _is to_ B _as_ R _is to_ X; this means that A × X = B × R. Supplying the values of the letters, we have 5 × 2 = 5 × 2. If we did not know the value of X, that is, if we were measuring the resistance of a coil of wire, using a 2-ohm coil as the standard, or R, we could find the value of X, knowing the other 3 parts of the proportion. 5 × X = 5 × 2, which means that 5 times the value of X is 10; hence the value of X is 10 ÷ 5 = 2 ohms. Suppose that we have R = 2 ohms, which is the standard resistance coil (No. 79), and are trying to find the resistance of a coil, X. We slide the end of wire, 3, along on the bridge-wire until the correct place is found. (See Exp. 125, 126, for details.) Take the values of A and B (§ 324), supply them in the equation given, and work out the value of X. =328. EXAMPLE.= R = 2 ohms; A = 3.7; B = 6.3; to find the value of X in ohms. A: B:: R: X, which means that A × X = B × R, or 3.7 × X = 6.3 × 2. X must equal, then (6.3 × 2) ÷ 3.7 = 3.405 ohms. =Note.= In practice it is most convenient to make connections as shown in Fig. 105 when measuring resistances (Exp. 126). The arrangement given in Fig. 102 is simply for explanation. It will be seen that the smaller A is, compared with B, the larger the unknown resistance compared with your standard. [Illustration: Fig. 105.] =EXPERIMENT 126. To measure the resistance of a wire by means of Wheatstone's Bridge; the "bridge method."=