325. Directions. (A) Touch the free end of wire, 3, to the point, C, which has a higher potential than M P. Press down K for an instant only. Some current should pass through A G, as a shunt. Should it pass from C to M P or the reverse? Note in which direction the right-hand end of the astatic needle is deflected.
(B) Swing the end of 3 around and touch it to the point, Z, which has a lower potential than M P. Press K for an instant, watch the needle, and compare with the results in (A).
(C) Move the free end of 3 along on G-s W, touching K at intervals, until a point is found at which the needle of A G is not deflected. How does the potential of this point compare with that of M P?
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[132] 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.