Sensitivity Of Wheatstone Bridge

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Sensitivity of Wheatstone’s bridge

The unknown resistance r is connected in the arm CD. Let the resistances in the arms AB, BC and AD be nmr1, mr1 and nr1 respectively. Let G be the galvanometer resitance. Let I be the current through the unknown resisitance r, Ig the current through the galvanometer and Ig the current through resistance nmr1. Then, by Kirchhoff’s I law, the currents through mr1 and nr1 are respectively (I1 + Ig) and (I + Ig).

When r = r1 the bridge is balanced and Ig = 0. Therefore, the difference (r – r1) is a measure of the want of balance. That is, the greater the difference (r – r1), the far is the bridge from balance. The bridge will be most sensitive, if the galvanometer current Ig is large, even when (r – r1) is extremely small.

Applying KVL to the loops ABDA and BCDB respectively,
            nmr1 I1 – GIg – nr1 (I + Ig) = 0        …… (1)

mr1 (I1 + Ig) – rI + GIg = 0            …… (2)

Multiplying Eq. (2) by n, we get
            nmr1I1 + nmr1Ig – nrI + nGIg = 0        …… (3)

Substracting Eq. (1) from Eq. (3), we get

G (1 + n)Ig + nr1 (1+m) Ig – n(r – r1) I = O

or            Ig/I = _________r – r1_________
                      (1 + 1/n) G + r1 (1 + m)        ……. (4)

For a given want to balance (r – r1), the value of Ig will be large when,
(i) I, the current through the unknown resistance is large.
(ii) G, the galvanometer resistance is small.
(iii) n, the ration P/Q = nmr1/mr1 is large.
(iv) m, the ration p/R = nmr1/nr1 is small.

The current I cannot be increased indefinitely because this will heat up the resistance coils and change their resistances. The sensitivity of the galvanometer itself is decreased if the resistance of the galvanotmeter G is too low.

Therefore, we have to increase the sensitivity by increasing n and decreasing m. In the ideal case, if n = ∞ and m = 0.
        (Ig/I)max = (r – r1) / (G + r1)

But this maximum sensitivity can never be realized in practice, because when n = ∞ and m = o, the points A and C are short circuited.
If m = n = 1
        Ig / I = (r – r1) / 2 (G + r1)

This value is still half of the value in the ideal case. In practice, this is the condition for maximum sensitivity. However, for good sensitiveness of the bridge, n > 1 and m < 1. Thus the Callendar rule is: The sensitivity of the bridge will be higher if the resistance in series with the unknown resistance is greater than the resistance connected in parallel to it.

The positions of the galvanometer and the battery can be interchanged but these two arrangements are not equally sensitive. Maxwell gave the following rule regarding the best arrangement of these.

Out of the battery and the galvanometer, the one having the higher resistance should be connected between the junction of the two highest resistances and the junction of the two lowest resistances.

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