3.14 Wheatstone Bridge

NCERT books for blind and low vision students.

As an application of Kirchhoffs rules consider the circuit shown in Fig. 3.25, which is called the Wheatstone bridge.

Figure 3.25 Weatstone Bridge

The bridge has four resistors R1, R2, R3 and R4. Across one pair of diagonally opposite points (A and C in the figure) a source of EMF is connected. This (i.e., AC) is called the battery arm. Between the other two vertices, B and D, a galvanometer G (which is a device to detect currents) is connected. This line, shown as BD in the figure, is called the galvanometer arm.

For simplicity, we assume that the cell has no internal resistance. In general there will be currents flowing across all the resistors as well as a current I g through G. Of special interest, is the case of a balanced bridge where the resistors are such that I g = 0. We can easily get the balance condition, such that there is no current through the galvanometer G. In this case, the Kirchhoffs junction rule applied to junctions D and B (see the figure) immediately gives us the relations I1 = I3 and I2 = I4. Next, we apply Kirchhoffs loop rule to closed loops ADBA and CBDC. The first loop gives

−I1 R1 + 0 + I2 R2 = 0 (since Ig = 0) .. .. (3.81)

and the second loop gives, upon using I3 = I1, I4 = I2


I2 R4 + 0 − I1 R3 = 0 .. .. (3.82)

From Equation (3.81), we obtain,


I1/I2 = R2/R1

whereas from Equation (3.82), we obtain,


I1/I2 = R4/R3

Hence, we obtain the condition


R2/R1 = R4/R3 .. .. [3.83(a)]

This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.

The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; R4 is thus not known. Keeping known resistances R1 and R2 in the first and second arm of the bridge, we go on varying R3 till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R4 is given by,


R4 = R3(R2/R1) .. .. [3.83(b)]

A practical device using this principle is called the meter bridge. It will be discussed in the next section.

Example 3.8

The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances:


AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω.

Figure 3.26:

A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Solution:


Considering the mesh BADB, we have


100I1 + 15Ig − 60I2 = 0


or 20I1 + 3Ig − 12I2= 0 .. .. [3.84(a)]

Considering the mesh BCDB, we have


10(I1 − Ig) − 15Ig − 5(I2 + Ig) = 0


10I1 − 30Ig − 5I2 = 0


2I1 − 6Ig − I2 = 0 .. .. [3.84(b)]

Considering the mesh ADCEA,


60I2 + 5(I2 + Ig) = 10


65I2 + 5Ig = 10


13I2 + Ig = 2 .. .. [3.84(c)]

Multiplying Equation (3.84b) by 10

20I1 − 60Ig − 10I2 = 0 .. .. [3.84(d)]

From Equations. (3.84d) and (3.84a) we have


63Ig − 2I2 = 0


I2 = 31.5Ig .. .. [3.84(e)]

Substituting the value of I2 into Equation [3.84(c)], we get


13 (31.5Ig ) + Ig = 2


410.5 Ig = 2


Ig = 4.87 mA.