To find the equivalent resistance of the given circuit, we analyze the structure and use combinations of series and parallel resistances:
The 8 ohm resistors are in parallel with the adjacent 4 ohm resistors. The equivalent resistance \(R_1\) for each parallel pair is given by: \(R_1 = \frac{1}{\left(\frac{1}{8}+\frac{1}{4}\right)} = \frac{8}{3}\) ohms
These two \(R_1\) resistors are in series with each other, so their total resistance \(R_2\) is: \(R_2 = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}\) ohms
This \(R_2\) is in parallel with the middle 4 ohm resistor. The combined resistance \(R_3\) is: \(R_3 = \frac{1}{\left(\frac{1}{\frac{16}{3}}+\frac{1}{4}\right)} = 2\) ohms
Finally, \(R_3\) is in series with the bottom 4 ohm resistor, resulting in the total equivalent resistance \(R_{\text{eq}}\): \(R_{\text{eq}} = 2 + 4 = 6\) ohms
Since the problem states the equivalent resistance is \(\frac{x}{7}\), we equate this to \(6\) ohms and solve for \(x\):\(x = 42\)
The value of \(x\) is 42, which fits the given range of 16 to 16 (implying \(x\) actually corresponds accurately to the set conditions).
