Question

A resistor of 8 ohms is connected in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the value of the resistor X?

Answer 1

When the two resistors are connected in parallel, we have:
$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{4.8}=\frac{1}{8}+\frac{1}{x} \\ \frac{1}{4.8}-\frac{1}{8}=\frac{1}{x} \\ \frac{1-0.6}{4.8}=\frac{1}{x} \\ \frac{0.4}{4.8}=\frac{1}{x} \\ x=12\mathrm{\Omega } \end{gathered}$$

x = 12 Ω
Therefore, the value of x = 12 Ώ