Question

From Ohm’s law for circuits,

it follows that the total resistance $R_{tot}$of two components connected in parallel is given by the equation

$R_{tot}=\frac{R_1{R}_2}{R_1+R_2}$

where $R_1,R_2$are individual resistance.

(a) Let $R_1=10\Omega$, and graph $R_{tot}$as a function of $R_2$.

(b) Find and interpret any asymptotes of the graph obtained in part (a).

(c) If $R_2=2\sqrt{R_1}$, what value of $R_1$will yield an $R_{tot}$of

17 ohms?

Short answer

(a)

(b) $R_{tot}=10$is the horizontal asymptote

(c) Forrole="math" localid="1646205899768" $R_{tot}=17\Omega required{R}_{1}willbe103.4\Omega$

Step-by-step solution

Part(a) Step 1. Given

$R_{tot}$of two components connected in parallel is given by the equation

$R_{tot}=\frac{R_1{R}_2}{R_1+R_2}$.

Part (a) Step 2. Calculation

Substituting the values we get:

$$\begin{gathered} R_{tot}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{10R_2}{10+R_2} \end{gathered}$$

putting this equation on graph we get

Part (b) Step 1. Given

$R_{tot}$of two components connected in parallel is given by the equation

$R_{tot}=\frac{R_1{R}_2}{R_1+R_2}$.

Part (b) Step 2. Calculation

To find the asymptote we solve the $R_{tot}$equation i.e.

$$\begin{gathered} R_{tot}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{10R_2}{10+R_2} \\ Taking{R}_{2}commonweget \\ =li{m}_{R_2\to \infty }\frac{10R_2}{R_2({\displaystyle \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$R_2+1)$}\right.}} \\ =10 \end{gathered}$$

Part (c) Step 1. Given

$R_{tot}$of two components connected in parallel is given by the equation

$R_{tot}=\frac{R_1{R}_2}{R_1+R_2}$.

Part (c) Step 2. Calculation

For $$\begin{gathered} R_2=2\sqrt{R_1} \\ substitutinginequation \\ {R}_{tot}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{2\sqrt{R_1}{R}_1}{R_1+\sqrt{2}{R}_1} \\ {R}_{tot}=17ohm \\ thereforeu\sin gthegraphfortheequation \\ 17=\frac{2\sqrt{R_1}{R}_1}{R_1+\sqrt{2}{R}_1} \\ weseethevaluefor{R}_1=103.4ohm \end{gathered}$$