Obtain the condition for bridge balance in Wheatstone's bridge.

Explanation

An important application of Kirchhoff.s rules is the Wheatstone.s bridge.
The bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is lo and its resistance is G.
Applying Kirchhoff.s current rule to junction B
` I_1 - I_G - I_3 = 0" ...(1)"`
Applying Kirchhoff.s current rule to junction D,
` I_2 + I_G - I_4 = 0" ...(2)"`

Applying Kirchhoff.s voltage rule to loop ABDA,
`I_1 P + I_G ""G - I_2 R = 0 " ...(2)"`
Applying Kirchhoff.s voltage rule to loop ABCDA,
`I_1 P+ I_3 Q - I_4 S - I_2 R = 0" ...(4)"`
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (`1G = 0`). Substituting lg = 0 in equation (1), (2) and (3), we get
`I_1 = I_3 " (5)"`
` I_2 = I_4 " ...(6)"`
`I_1 P = I_2 R " ...(7)"`
Substituting the equation (5) and (6) in equation (4)
`I_1 P + I_1 Q - I_2 S - I_2 R = 0`
`I_2 (P+Q) = I_2 (R+S)" ...(8)"`
Dividing equation (8) by equation (7), we get
`(P+Q)/P = (R+S)/R`
` 1 + Q/P =1 + S/R`
`Q/P = S/R`
`P/Q = R/S" ...(9)"`
This is the bridge balance condition. Only under this condition, galvanometer shows null deflection.