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Obtain the balancing condition for the Wheatstone bridge arrangement as shown in Figure 4 below:

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Consider the diagram as shown below. The distribution of current in accordance with Kirchhoff’s rules is shown. Applying Kirchhoff’s loop rule to the closed loop ABDA, we have

-I1P - IgG+(I - I1)R = 0 ...(1)

Again, applying Kirchhoff's loop rule o closed loop BCDE, we have

-(I1 - Ig)Q + (I - I1 + Ig)X + IgG = 0 ...(2)

This value of P,O,R and X are so adjusted that the galvanometer gives zero deflection i.e., Ig = 0.

This situations is the balanced state of the Wheatstone bride. Putting Ig = 0 in equations (1) and (2), We have

-I1P + (I - I1)R = 0 .......(3)

and  -I1Q+(I - I1)X = 0 ......(4)

Rewriting the above two equations, we have

I1P = (I - I1)R  .....(5)

and   I1Q = (I - I1)X ....(6)

Divinding the above equations, we have

\(\frac{P}{Q} = \frac{R}{X}\) ...(7)

The above expression gives the condition for the bance of a wheatstone bridge.

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