The equivalent resistance between the points A and B in the following circuit is x/5Ω. The value of x is ______. Numerical answer type. Work it out, then open the answer.

Answer: 21. Left wire puts the bottom-left at A; right wire puts the top-right at B. So: A–M 6, M–B 3, A–N 3, N–B 6, M–N 3 Ω (unbalanced bridge). Take VA = 1, VB = 0. KCL at M: 5VM - 2VN = 1; at N: 5VN - 2VM = 2. So VM = 37, VN = 47. Current from A: (1 - 3/7)/(6) + (1 - 4/7)/(3) = 2/21 + 3/21 = 5/21 RAB = 21/5 Ω, so x = 21.

The equivalent resistance between the points A and B in the following… | JEE Main 2026 (28 Jan, Shift 1)

Q5JEE Main 2026 · 28 JAN · SHIFT 1Medium

Combination of Resistors

The equivalent resistance between the points $A$ and $B$ in the following circuit is $\frac{x}{5}\Omega$ . The value of $x$ is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 21

Left wire puts the bottom-left at A; right wire puts the top-right at B. So: A–M $6$ , M–B $3$ , A–N $3$ , N–B $6$ , M–N $3\ \Omega$ (unbalanced bridge).
Take $V_A = 1$ , $V_B = 0$ . KCL at M: $5V_M - 2V_N = 1$ ; at N: $5V_N - 2V_M = 2$ . So $V_M = \tfrac37$ , $V_N = \tfrac47$ .
Current from A: $\dfrac{1 - 3/7}{6} + \dfrac{1 - 4/7}{3} = \dfrac{2}{21} + \dfrac{3}{21} = \dfrac{5}{21}$
$R_{AB} = \dfrac{21}{5}\ \Omega$ , so $x = 21$ .

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