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+1 vote
27.9k views
in Physics by (1.4k points)
edited by

State Kirchhoff’s law of electrical network using these laws drive an expression for the balance condition in a Wheatstone bridge.

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2 Answers

+3 votes
by (323k points)

A Wheatstone bridge arrangement is shown as below:

Using Kirchoff's seconds law to the loop ABDA, we get

I1P - IgG - I2R = 0: G is the galvanometer resistance.

Applying Kirchoff's law to the loop ABDA, we get

(I1 - Ig)Q - (I2 - Ig)S - GIg = 0

When the bridge is balanced Ig = 0

Then, the equations can be written as,

IP - I2R = 0 or I1P = I2R .........(1)

I1Q - I2S = 0 or I1Q = I2S ..........(2)

On dividing equation (1) by (2), we get

P/Q = R/S, which is the balanced condition of a Wheatstone bridge.

0 votes
by (15 points)

Current arrangement for a wheatstone bridge is being shown in the adjoining fig. in balance condition of bridge no current flows.

Through galvanometer and it's gives no deflecation. currents in various branches of network as per kirchhoff's first law are,

Thus represented in figure.

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