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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.
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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 flows 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. The 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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