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Questions
With the help of a labelled diagram, show that the balancing condition of a Wheatstone bridge is
`"R"_1/"R"_2 = "R"_3/"R"_4` where the terms have their usual meaning.
Obtain the balancing condition in the case of Wheatstone's network.
Solution
Four resistances P, Q, R and S are connected to form a quadrilateral ABCD as shown in the following figure. A battery of emf ε along with a key is connected between points A and C such that point A is at higher potential with respect to point C. A galvanometer of internal resistance G is connected between points B and D.
When the key is closed, current I flow through the circuit. It divides into I1 and I2 at point A. I1 is the current through P and I2 is the current through S. The current I1 gets divided at point B. Let Ig be the current flowing through the galvanometer. The currents flowing through Q and R are respectively (I1 – Ig) and (I2 + Ig),
I = I1 + I2 ...(1)
Consider the loop ABDA. Applying Kirchhoff’s voltage law in the clockwise sense shown in the loop we get,
- I1P - IgG + I2S = 0 ...(2)
Applying Kirchhoff's voltage law to loop BCDB in a clockwise sense, we get,
- (I1 - Ig)Q + (I2 + Ig)R + IgG = 0 .....(3)
From these three equations (Eq. (1), (2), (3) we can find the current flowing through any branch of the circuit.
A special case occurs when the current passing through the galvanometer is zero. In this case, the bridge is said to be balanced. Condition for the balance is Ig = 0. This condition can be obtained by adjusting the values of P, Q, R and S. Substituting Ig = 0 in Eq. (2) and Eq. (3) we get,
– I1P + I2S = 0 ∴ I1P = I2S ...(4)
– I1Q + I2R = 0 ∴ I1Q = I2R ...(5)
Dividing Eq. (4) by Eq. (5), we get
`∴ ("I"_1"P")/("I"_1"Q") = ("I"_1"S")/("I"_2"R")`
`therefore"P"/"Q" = "S"/"R"`
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