The equivalent resistance between the points $A$ and $B$ in the following circuit is $\dfrac x5\Omega$. The value of $x$ is ______ .
Numerical value type. Enter your answer.
Answer: 21
Label the top middle junction $M$ and the bottom middle junction $N$. Then $A$–$M$: $6\ \Omega$, $M$–$B$: $3\ \Omega$, $A$–$N$: $3\ \Omega$, $N$–$B$: $6\ \Omega$, and $M$–$N$: $3\ \Omega$. The bridge is not balanced ($6/3 \ne 3/6$).
Take $V_A = 1$ V, $V_B = 0$ and apply KCL.
At $M$: $\dfrac{1 - V_M}{6} = \dfrac{V_M}{3} + \dfrac{V_M - V_N}{3}\Rightarrow 5V_M - 2V_N = 1$
At $N$: $\dfrac{1 - V_N}{3} = \dfrac{V_N}{6} + \dfrac{V_N - V_M}{3}\Rightarrow 5V_N - 2V_M = 2$
Solving: $V_M = \dfrac37$ V, $V_N = \dfrac47$ V.
Current leaving $A$:
$$I = \frac{1 - \frac37}{6} + \frac{1 - \frac47}{3} = \frac{2}{21} + \frac{3}{21} = \frac5{21}\ \text{A}$$
$$R_{AB} = \frac1I = \frac{21}{5}\ \Omega\Rightarrow x = 21$$
Solution by Sreeraj P, M.Sc Physics