Question

In Figure 19.76 the resistance ${\mathit{R}}_{\mathbf{1}}$ is $\mathbf{10}\mathbf{}\mathbf{\Omega }$, ${\mathit{R}}_{\mathbf{2}}$is$\mathbf{5}\mathbf{}\mathbf{\Omega }$ , and ${\mathit{R}}_{\mathbf{3}}$ is $\mathbf{20}\mathbf{}\mathbf{\Omega }$. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Short answer

$14\mathrm{\Omega }$

Step-by-step solution

Given Data

$$\begin{gathered} R_1=10\mathrm{\Omega } \\ {R}_2=5\mathrm{\Omega } \\ {R}_3=20\mathrm{\Omega } \end{gathered}$$

Concept

The equivalent resistance is the addition of the resistances, when the batteries are connected in series.

Determine the equivalent resistance

Resistors $R_2$and $R_3$are connected in parallel,

role="math" localid="1662186967334" $\begin{array}{rcl}\frac{1}{R}& =& \frac{1}{R_2}+\frac{1}{R_3}\\ & =& \frac{1}{5}+\frac{1}{20}\\ & =& 0.25\\ R_4& =& 1/0.25\\ & =& 4\mathrm{\Omega }\end{array}$

Resistors $R_1$and $R_4$are connected in series,

$\begin{array}{c}{R}_5=R_1+R_4\\ =10+4\\ =14\mathrm{\Omega }\end{array}$

Hence, the equivalent resistance is$14\mathrm{\Omega }$