Precalculus (6th Edition) Blitzer

The simplified form of the rational expression is $\frac{{{R}_{1}}{{R}_{2}}{{R}_{3}}}{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}}$. The combined resistance when ${{R}_{1}}=4\text{ ohms, }{{R}_{2}}=8\text{ ohms }{{R}_{3}}=12\text{ ohms}$ is $\frac{24}{11}\text{ohms}$.
Consider the expression, $\frac{1}{\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}}$ Therefore, \begin{align} & \frac{1}{\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}}=\frac{1}{\frac{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}}{{{R}_{1}}{{R}_{2}}{{R}_{3}}}} \\ & =\frac{{{R}_{1}}{{R}_{2}}{{R}_{3}}}{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}} \end{align} Thus, the simplified form of the provided expression is $\frac{{{R}_{1}}{{R}_{2}}{{R}_{3}}}{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}}$ Now, substitute the value ${{R}_{1}}=4\text{ ohms, }{{R}_{2}}=8\text{ ohms }{{R}_{3}}=12\text{ ohms}$ in the expression, $\frac{{{R}_{1}}{{R}_{2}}{{R}_{3}}}{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}}$ Therefore, \begin{align} & \frac{{{R}_{1}}{{R}_{2}}{{R}_{3}}}{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}}=\frac{4\cdot 8\cdot 12}{8\cdot 12+4\cdot 12+4\cdot 8} \\ & =\frac{384}{176} \\ & =\frac{24}{11} \end{align} The simplified form of the rational expression is $\frac{{{R}_{1}}{{R}_{2}}{{R}_{3}}}{{{R}_{2}}{{R}_{3}}+{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{2}}}$. The combined resistance when ${{R}_{1}}=4\text{ ohms, }{{R}_{2}}=8\text{ ohms }{{R}_{3}}=12\text{ ohms}$ is $\frac{24}{11}\text{ohms}$.