CBSE Class 12-science Answered
How to solve the Wheatstone bridge.
Asked by kripanjalihimansu | 28 Feb, 2019, 05:56: AM
Expert Answer
Figure shows 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. Between the other two vertices,
B and D, a galvanometer G is connected. 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 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)
and the second loop gives, upon using I3 = I1, I4 = I2
and the second loop gives, upon using I3 = I1, I4 = I2
I2 R4 + 0 – I1 R3 = 0 ..................................(2)
From Eq. (1), we obtain,
whereas from Eq. (2), we obtain,
Hence, we obtain the condition
This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.
Answered by Thiyagarajan K | 28 Feb, 2019, 12:39: PM
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