WHEATSTONE’S BRIDGE

If the current in OA is zero, the potential at O must be equal to that at A. Now when we know the potentials at B and C we can determine those at O and A by the rule given at Art. 157, provided there is no current in OA,

O= Bγ + Cβ , β+γ A= Bb + Cc , b+c Fig. 48. whence the condition is bβ = cγ, where b, c, β, γ are the resistances in CA, AB, BO and OC respectively. To determine the degree of accuracy attainable by this method we must ascertain the strength of the current in OA when this condition is not fulfilled exactly.

Let A, B, C and O be the four points. Let the currents along BC, CA and AB be x, y and z, and the resistances of these conductors a, b and c. Let the currents along OA, OB and OC be ξ, η, ζ, and the resistances α, β and γ. Let an electromotive force E act along BC. Required the current ξ along OA. Let the potentials at the points A, B, C and O be denoted by the symbols A, B, C and O. The equations of conduction are ax = B − C + E, by = C − A cz = A − B αξ = O − A, βη = O − B, γζ = O − C; with the equations of continuity ξ + y − z = 0, η + z − x = 0, ζ + x − y = 0. By considering the system as made up of three circuits OBC, OCA and OAB in which the currents are x, y, z respectively, and applying Kirchhoff’sWHEATSTONE’S BRIDGE. 209 rule [Art. 158] to each cycle, we eliminate the values of the potentials O, A, B, C, and the currents ξ, η, ζ, and obtain the following equations for x, y and z, (a + β + γ)x − γy − βz = E, −γx + (b + δ + α)y − αz = 0, −βx − αy + (c + α + β)z = 0. Hence, if we put −γ −β | |α + β + γ b+γ +α −α | , D = | −γ | | | −β −α c + α + β| we find ξ=E (bβ − cγ), D x=E {(b + γ)(c + β) + α(b + c + β + γ)}. D and 222*.] The value of D may be expressed in the symmetrical form, D = abc + bc(β + γ) + ca(γ + α) + ab(α + β) + (a + b + c)(βγ + γα + αβ) or, since we suppose the battery in the conductor a and the galvanometer in α, we may put B the battery resistance for a and G the galvanometer resistance for α. We then find D = BG(b + c + β + γ) + B(b + γ)(c + β)

• G(b + c)(β + γ) + bc(β + γ) + βγ(b + c). If the electromotive force E were made to act along OA, the resistance of OA being still α, and if the galvanometer were placed in BC, the resistance of BC being still a, then the value of D would remain the same, and theWHEATSTONE’S BRIDGE.

current in BC due to the electromotive force E acting along OA would be equal to the current in OA due to the electromotive force E acting in BC. But if we simply disconnect the battery and the galvanometer, and without altering their respective resistances connect the battery to O and A and the galvanometer to B and C, then in the value of D we must exchange the values of B and G. If D′ be the value of D after this exchange, we find D − D′ = (G − B) {(b + c)(β + γ) − (b + γ)(β + c)} , = (B − G) {(b − β)(c − γ)} . Let us suppose that the resistance of the galvanometer is greater than that of the battery.

Let us also suppose that in its original position the galvanometer connects the junction of the two conductors of least resistance β, γ with the junction of the two conductors of greatest resistance b, c, or, in other words, we shall suppose that if the quantities b, c, γ, β are arranged in order of magnitude, b and c stand together, and γ and β stand together. Hence the quantities b − β and c − γ are of the same sign, so that their product is positive, and therefore D − D′ is of the same sign as B − G.

If therefore the galvanometer is made to connect the junction of the two greatest resistances with that of the two least, and if the galvanometer resis- tance is greater than that of the battery, then the value of D will be less, and the value of the deflexion of the galvanometer greater, than if the connexions are exchanged.

The rule therefore for obtaining the greatest galvanometer deflexion in a given system is as follows: Of the two resistances, that of the battery and that of the galvanometer, connect the greater resistance so as to join the two greatest to the two least of the four other resistances. 223*.] We shall suppose that we have to determine the ratio of the resis- tances of the conductors AB and AC, and that this is to be done by finding a point O on the conductor BOC, such that when the points A and O are connected by a wire, in the course of which a galvanometer is inserted, no sensible deflexion of the galvanometer needle occurs when the battery is made to act between B and C.WHEATSTONE’S BRIDGE.

The conductor BOC may be supposed to be a wire of uniform resistance divided into equal parts, so that the ratio of the resistances of BO and OC may be read off at once.

Instead of the whole conductor being a uniform wire, we may make the part near O of such a wire, and the parts on each side may be coils of any form, the resistance of which is accurately known. We shall now use a different notation instead of the symmetrical notation with which we commenced.

Let the whole resistance of BAC be R. Let c = mR and b = (1 − m)R. Let the whole resistance of BOC be S. Let β = nS and γ = (1 − n)S. The value of n is read off directly, and that of m is deduced from it when there is no sensible deviation of the galvanometer. Let the resistance of the battery and its connexions be B, and that of the galvanometer and its connexions G. We find as before D = G{BR + BS + RS} + m(1 − m)R2 (B + S) + n(1 − n)S 2 (B + R)

• (m + n − 2mn)BRS, and if ξ is the current in the galvanometer wire ξ= ERS (n − m). D

In order to obtain the most accurate results we must make the deviation of the needle as great as possible compared with the value of (n − m). This may be done by properly choosing the dimensions of the galvanometer and the standard resistance wire.

WHEATSTONE’S BRIDGE

It may be shewn that when the form of a galvanometer wire is changed while its mass remains constant, the deviation of the needle for unit current is proportional to the length, but the resistance increases as the square of the length. Hence the maximum deflexion is shewn to occur when the resistance

of the galvanometer wire is equal to the constant resistance of the rest of the circuit. In the present case, if δ is the deviation, δ = C√Gξ, where C is some constant, and G is the galvanometer resistance which varies as the square of the length of the wire. Hence we find that in the value of D, when δ is a maximum, the part involving G must be made equal to the rest of the expression.

If we also put m = n, as is the case if we have made a correct observation, we find the best value of G to be G = n(1 − n)(R + S). This result is easily obtained by considering the resistance from A to O through the system, remembering that BC, being conjugate to AO, has no effect on this resistance.

In the same way we should find that if the total area of the acting surfaces of the battery is given, the most advantageous arrangement of the battery is when RS B= . R+S Finally, we shall determine the value of S such that a given change in the value of n may produce the greatest galvanometer deflexion. By differentiat- ing the expression for ξ we find S2 = BR G R+ . B+R( n(1 − n) ) If we have a great many determinations of resistance to make in which the actual resistance has nearly the same value, then it may be worth while to prepare a galvanometer and a battery for this purpose. In this case we find that the best arrangement is S = R, and if n = 12 , G = 12 R. B = 12 R, G = 2n(1 − n)R,USE OF WHEATSTONE’S BRIDGE.