Question

When resistors 1 and 2 are connected in series, the equivalent resistance is 16.0 Ω. When they are connected in parallel, the equivalent resistance is 3.0 Ω. What are the smaller resistance and the larger resistance of these two resistors?

Short answer

The smaller resistance is ${\mathrm{R}}_1=4\mathrm{\Omega }$and the larger resistance is${\mathrm{R}}_2=12\mathrm{\Omega }$

Step-by-step solution

Given

1. The equivalent resistance of the series arrangement is${\mathrm{R}}_{\mathrm{eq}}=16\mathrm{\Omega }$
2. The equivalent resistance of the parallel arrangement is ${\mathrm{R}}_{\mathrm{eq}}=3\mathrm{\Omega }$

Determining the concept

Use the concept of equivalent resistance for series circuit and parallel circuit. Using the equations, find the larger and the smaller resistance.

Formulae are as follow:

Resistance in series,

${\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_1+{\mathrm{R}}_2$

Resistance in parallel,

$\frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}$

Determining the smaller and larger resistance.

The two equations can be written as ,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_1+{\mathrm{R}}_2 \\ 16={\mathrm{R}}_1+{\mathrm{R}}_2.......................\left(1\right) \\ \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \end{gathered}$$

Write it as,

$$\begin{gathered} \mathrm{R}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \\ 3.0\mathrm{\Omega }=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}.............................\left(2\right) \end{gathered}$$

Using the value of ${\mathrm{R}}_1+{\mathrm{R}}_2$from equation (1) in equation (2),

$$\begin{gathered} 3.0=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{16} \\ \mathrm{RR}=3.0\times 16=48 \end{gathered}$$

From the equation (1), it can be determined that ${\mathrm{R}}_1=4\mathrm{\Omega }$and ${\mathrm{R}}_2=12\mathrm{\Omega }$

These values also satisfy the equation (2). So, the smaller resistance is ${\mathrm{R}}_1=4\mathrm{\Omega }$and the larger resistance is${\mathrm{R}}_2=12\mathrm{\Omega }$

Hence, the smaller resistance is ${\mathrm{R}}_1=4\mathrm{\Omega }$and the larger resistance is${\mathrm{R}}_2=12\mathrm{\Omega }$

Therefore, by using the concept of equivalent resistance for series circuit and parallel circuit smaller and larger resistance can be determined.