Electrical measuring instruments

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Question

Four wires of equal length 8 m are arranged as shown in the figure. Wires, 2, 3 and 4 are of equal cross sectional area and wire 1 is of half the cross section of these wires. By how much distance $(i n m e t r e)$ pointer at point 5 must be moved to get null point?

Moderate

Solution

For getting null point
$\begin{array}{l} \frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}} \\ as R 2 = R 4 \\ R 1 = R 3 \end{array}$
Let the pointer at point 5 is moved to left by distance x to get nul point as shown in the figure. If resistance per unit length of wire 3 is r then that of wire 1 will be 2r.
$\begin{array}{l} (8 - x) 2 r = x \times 2 r + 8 \times r \\ 4 x = 8 \\ x = 2 m \end{array}$

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