91Ó°ÊÓ

1. The equivalent resistance of the series arrangement is$R_{eq}=16\text{Ω}$
2. The equivalent resistance of the parallel arrangement is$R_{eq}=3.0\text{Ω}$

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,

$R_{eq}=R_1+R_2$

Resistance in parallel,

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

The two equations can be written as ,

$$\begin{gathered} R_{eq}=R_1+R_2 \\ 16=R_1+R_2……………………………………………………………………………….…\mathrm{..}\left(1\right) \end{gathered}$$

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Write it as,

$$\begin{gathered} R_{eq}=\frac{R_1{R}_2}{R_1+R_2} \\ 3.0=\frac{R_1{R}_2}{R_1+R_2}……………………………………………………………………………………\left(2\right) \end{gathered}$$

Using the value of$R_1+R_2$from equation (1) in equation (2),

$$\begin{gathered} 3.0=\frac{R_1{R}_2}{16} \\ {R}_1{R}_2=3.0×16=48 \end{gathered}$$

From the equation (1), it can be determined that$R_1=4\text{Ω}$ and$R_2=12\text{Ω}$

These values also satisfy the equation (2). So, the smaller resistance is $R_1=4\text{Ω}$and the larger resistance is$R_2=12\text{Ω}$

Hence, the smaller resistance is$R_1=4\text{Ω}$and the larger resistance is$R_2=12\text{Ω}$

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