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Write Kirchhoff's laws and establish the necessary balance condition for Wheatstone's bridge
Asked by divyabharti14042004 | 18 Jan, 2022, 02:52: PM
(a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the
sum of currents leaving the junction

(b) Loop rule: The algebraic sum of changes in potential around any closed loop involving
resistors and cells in the loop is zero

As an application of Kirchhoff’s rules consider the circuit shown in above figure, which is called the Wheatstone bridge.
The bridge has four resistors R1, R2, R3 and R4. Across one pair of diagonally opposite
points (A and C in the figure) a source is connected. This (i.e., AC) is called the battery arm.
Between the other two vertices, B and D, a galvanometer G (which is a device to detect currents) is connected.
This line, shown as BD in the figure, is called the galvanometer arm.

For simplicity, we assume that the cell has no internal resistance. In general there will be currents flowing across
all the resistors as well as a current Ig through G. Of special interest, is the case of a balanced bridge
where the resistors are such that Ig = 0. We can easily get the balance condition, such that there is no current through G.

In this case, the Kirchhoff’s junction rule applied to junctions D and B (see the figure) immediately gives us the
relations I1 = I3 and I2 = I4.

Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives

–I1 R1 + 0 + I2 R2 = 0....................................(1)

Current through galavanometer is zero, Hence middle term zero in above eqn.(1)

Second loop gives, upon using I3 = I1, I4 = I2

I2 R4 + 0 – I1 R3 = 0  .....................................(2)

From Eqn. (1), we obtain,

whereas from Eqn. (2), we obtain,

Hence, we obtain the condition

This last expression relating the four resistors is called the balance condition for the
galvanometer to give zero or null deflection
Answered by Thiyagarajan K | 18 Jan, 2022, 05:47: PM

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