Finding Resistance of an Individual Resistor

Finding the Resistance of an Individual Resistor

Example 1

Given that the equivalent resistance of the connection in the figure above is 0.8Ω, find the resistance of the resistor R.

Answer:
$$\begin{array}{l}R={\left(\frac{1}{2}+\frac{1}{R}+\frac{1}{4}\right)}^{-1}=0.8\\ \left(\frac{1}{2}+\frac{1}{R}+\frac{1}{4}\right)={0.8}^{-1}=\frac{5}{4}\\ \frac{1}{R}=\frac{5}{4}-\frac{1}{2}-\frac{1}{4}=\frac{1}{2}\\ R=2\Omega \end{array}$$

Example 2

Given that the equivalent resistance of the connection in the figure above is 4Ω, find the resistance of the resistor R.

Answer:
$$\begin{array}{l}{\left(\frac{1}{2+R}+\frac{1}{3}\right)}^{-1}+2=4\\ {\left(\frac{1}{2+R}+\frac{1}{3}\right)}^{-1}=4-2=2\\ \left(\frac{1}{2+R}+\frac{1}{3}\right)=2^{-1}\\ \frac{1}{2+R}=\frac{1}{2}-\frac{1}{3}\\ \frac{1}{2+R}=\frac{1}{6}\\ 2+R=6\\ R=4\Omega \end{array}$$