A wire of uniform cross-section and resistance 4 $\Omega$ is bent i...

Question

A wire of uniform cross-section and resistance 4 $\Omega$ is bent into a square ABCD. Point A is connected to point P on DC by a wire AP of resistance 1 $\Omega$. When potential difference is applied between A and C, points B and P are at same potential. Find resistance of part DP.

Accepted Answer

or

Each side of square = 1 $\Omega$. Let $DP = x\ \Omega$, then $PC = (1-x)\ \Omega$.

Since B and P are at same potential, using Wheatstone balance:

$$\frac{AB}{BC} = \frac{AP}{PC} \Rightarrow \frac{1}{1} = \frac{1}{1-x}$$

Solving: $DP = 0.5\ \Omega$

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