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Question
State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.
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Solution
Kirchhoff’s First Law − Junction Rule
The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
Let the currents be I1, I2I3, and I4
Convention:
Current towards the junction − positive
Current away from the junction − negative
I3 + (− I1) + (− I2) + (− I4) = 0
Kirchhoff’s Second Law − Loop Rule
In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.
For closed part BACB, E1 − E2 = I1R1 + I2 R2 − I3R3
For closed part CADC, E2 = I3R3 + I4R4 + I5R5
Wheatstone Bridge:
The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure.
R1, R2, R3,and R4 are the four resistances.
Galvanometer (G) has a current Ig flowing through it at balanced condition, Ig = 0
Applying junction rule at B,
∴ I2 = I4
Applying junction rule at D,
∴ I1 = I3
Applying loop rule to closed loop ADBA,
`-I_1R_1+0+I_2R_2 = 0`
`∴ I_1/I_2 = R_2/R_1 ...... (1)`
Applying loop rule to closed loop CBDC,
`I_2R_4+0-I_1R_3 = 0` ∵`I_3 =I_1,I_4 =I_2`
`∴ I_1/I_2 = R_4/R_3 ...... (2)`
From equations (1) and (2),
`R_2/R_1 = R_4/R_3`
This is the required balanced condition of Wheatstone Bridge.
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