In the circuit shown below in figure, calculate the value of x if the equivalent resistance between the points A and B is 4 Ω.

In the circuit, there are three parts. In the first part, resistors of 4 Ω and 8 Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = 4 + 8 = 12 Ω

In the second part, resistors of x Ω and 5 Ω are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (x + 5) Ω

In the third part, the two parts of resistance R'_s = 12 Ω and R''_s = (x + 5) Ω are connected in parallel. If the equivalent resistance between points A and B is R_p then

$\dfrac{1}{R$p} = \dfrac{1}{R's} + \dfrac{1}{R''s} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{12} + \dfrac{1}{x + 5} \\[0.5em]

But it is given that equivalent resistance between points A and B is 4 Ω
∴ R_p = 4 Ω

Putting value of R_p in the equation above and solving for x:

$\dfrac{1}{4} = \dfrac{1}{12} + \dfrac{1}{x + 5} \\[0.5em] \dfrac{1}{4} = \dfrac{x + 5 +12}{12(x+5)} \\[0.5em] \dfrac{1}{4} = \dfrac{x + 17}{12(x+5)} \\[0.5em] \dfrac{12(x+5)}{4} = x + 17 \\[0.5em] 3(x+5) = x + 17 \\[0.5em] 3x + 15 = x + 17 \\[0.5em] 3x - x = 17 - 15 \\[0.5em] 2x = 2 \\[0.5em] \Rightarrow x = 1 Ω$

Hence, the value of x = 1 Ω