91影视

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

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,

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

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,

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

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

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

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

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

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

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