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Explanation, Formulas, Solved Example Problems | Kirchhoff’s rule - Wheatstone’s bridge | 12th Physics : Current Electricity

Chapter: 12th Physics : Current Electricity

Wheatstone’s bridge

An important application of Kirchhoff’s rules is the Wheatstone’s bridge.

Wheatstone’s bridge

An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected as shown in Figure 2.25. 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 IG and its resistance is G.

Applying Kirchhoff’s current rule to junction B


Applying  Kirchhoff’s  current  rule  to junction D,



Applying Kirchhoff’s voltage rule to loop ABDA,


Applying Kirchhoff’s voltage rule to loop ABCDA,


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 (IG = 0). Substituting IG = 0 in equation (2.45), (2.46) and (2.47), we get


Substituting  the  equation  (2.49)  and (2.50) in equation (2.48)    


Dividing equation (2.52) by equation (2.51), we get


This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

 

EXAMPLE 2.23

In a Wheatstone’s bridge P = 100 Ω, Q = 1000 Ω and R = 40 Ω. If the galvanometer shows zero deflection, determine the value of S.

Solution


 

EXAMPLE 2.24

What is the value of x when the Wheatstone’s network is balanced?

P = 500 Ω, Q = 800 Ω, R = x + 400, S = 1000 Ω


Solution

P/Q = R/S


x + 400 = 0.625 × 1000

x + 400 = 625

x = 625 – 400

x = 225 Ω

 

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