If the potential difference between B and D is zero, the value of… — Solution | JEE Physics

NVAMediumJEE 2023Kirchhoff's Laws & Circuits

If the potential difference between B and D is zero, the value of $x$ is $\frac{1}{n} \, \Omega$ . The value of $n$ is ........

Answer

Correct answer:2

Standard Method

Given: The potential difference between B and D is zero.

Find: The value of $n$ if $x = \frac{1}{n} \, \Omega$ .

For points B and D to be at the same potential in this resistive network, the voltage division along the two branches must balance.

Using the relation given in the solution:

\frac{2}{3} = \frac{x+1}{x}

Now solve for $x$ :

2x = 3(x + 1)

2x = 3x + 3

x = 0.5 = \frac{1}{2}

Since $x = \frac{1}{n} \, \Omega$ , we get

\frac{1}{n} = \frac{1}{2}

Therefore, $n = 2$ .

Thus, the required numerical value is $2$ .

Using the condition of zero current instead of equal potential at B and D is incorrect. Zero potential difference means the two nodes are equipotential, so the branch voltage drops must balance. Use the balancing condition between the resistor ratios.
Solving $\frac{2}{3} = \frac{x+1}{x}$ incorrectly by cross-multiplication can lead to a wrong value of $x$ . Multiply carefully first, then isolate $x$ step by step.
Reporting $x = \frac{1}{2} \, \Omega$ as the final answer is wrong because the question asks for $n$ , not $x$ . After finding $x = \frac{1}{2}$ , compare with $x = \frac{1}{n}$ to obtain $n = 2$ .

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