Hawkins Electrical Guide v. 03 (of 10): Questions, Answers, & Illustrations, A progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applicationsHawkins, N. (Nehemiah)
Science
Hawkins Electrical Guide v. 03 (of 10): Questions, Answers, & Illustrations, A progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications
Hawkins, N. (Nehemiah)
Electrical engineering -- Handbooks, manuals, etc.
The bridge consists of a system of conductors as shown in fig. 564. The
circuit of a constant battery is made to branch at P into two parts, which
re-unite at Q, so that part of the current flows through the point M,
the other part through the point N. The four conductors A, B, C, D, are
spoken of as the _arms_ of the balance or bridge. It is by the proportion
existing between the resistances of these arms that the resistance of
one of them can be calculated when the resistances of the other three
are known. When the current which starts from the battery arrives at P,
the pressure will have fallen to a certain value. The pressure in the
upper branch falls again to M, and continues to fall to Q. The pressure
of the lower branch falls to N, and again falls till it reaches the value
at Q. Now if N be the same proportionate distance along the resistances
between P and Q, as M is along the resistances of the upper line between
P and Q, the pressure will have fallen at N to the same value as it has
fallen to at M; or, in other words, if the ratio of the resistance C to
the resistance D be equal to the ratio between the resistance A and the
resistance B, then M and N will be at equal pressures. To find out if
this condition obtain, a sensitive galvanometer is placed in a branch wire
between M and N which will show _no_ deflection when M and N are at equal
pressure or when the four resistances of the arms "balance" one another
by being in proportion, thus:
(1) A:C = B:D
If, then, the value of A, B, and C be known, D can be calculated. The
proportion (1) is reduced to the following equation before substituting.
D = BC/A
For instance, if A and C be, as in fig. 565, 10 ohms and 100 ohms
respectively, and B be 15 ohms, D will be (15 × 100) ÷ 10 = 150 ohms.
[Illustration: Fig. 566.--Diagram showing usual arrangement of resistances
in arms of Wheatstone's bridge. In practice the bridge is seldom or never
made in the lozenge shape of the diagrams, figs. 564 and 565, these being
given merely for clearness. The resistance box of fig. 554 is, in itself,
a complete "bridge," the appropriate connections being made by screws at
various points. The letters in the above diagram correspond with those in
figs. 564 and 565, and the three figures should be carefully compared.]
As constructed, Wheatstone bridges are provided with some resistance
coils in the arms A and C, as well as with a complete set in the arm
B. The advantage of this arrangement is that by adjusting A and C, the
proportionality between B and D can be determined, and can, in certain
cases, be measured to fractions of an ohm. In fig. 565 resistances of 10,
100, and 1,000 ohms are included in the arms A and C.
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