Question

The equivalent resistance of a series combination of two resistors is $s$. When they are connected in parallel, the equivalent resistance is $p$. If $s=np$, then the maximum value for $n$ is ______. (Round off to the nearest integer)

Answer 1

Step 1: Given data

The equivalent resistance of a series combination is $s$.

The equivalent resistance of a parallel combination is $p$

The relationship between the equivalent resistance of series and parallel combinations is

$s=np$

Step 2: Formula used

Suppose the two are resistors are $R_1$ and $R_2$.

The equivalent resistance of two resistors $R_1$ and $R_2$ when they are in series combination are-

$s=R_1+R_2$

When parallel connected then-

$$\begin{gathered} \frac{1}{p}=\frac{1}{R_1}+\frac{1}{R_2} \\ p=\frac{R_1{R}_2}{R_1+R_2} \end{gathered}$$

Step 3: Compute the maximum value for n

Substitute the known values in the relation $s=np$,

$R_1+R_2=n\times \frac{R_1{R}_2}{R_1+R_2}$

Multiply both sides by$\left(R_1+R_2\right)$,

$$\begin{gathered} {\left(R_1+R_2\right)}^2=n{R}_1{R}_2 \\ {R}_1^2+R_2^2+\left(2-n\right)R_1{R}_2=0 \end{gathered}$$

Divide the whole equation by ${R_2}^2$

$\frac{{R_1}^2}{{R_2}^2}+1+\left(2-n\right)\frac{R_1}{R_2}$

Let $\frac{R_1}{R_2}=k$

Now simplifying further,

$$\begin{gathered} k^2+1+\left(2-n\right)k=0 \\ {k}^2+\left(2-n\right)k+1=0......\left(1\right) \end{gathered}$$

On comparing equation (1) with $a{x}^2+bx+c=0$

$a=1,b=\left(2-n\right),c=1$

Use the Dharacharya rule to compute a discriminant,

$D\ge 0$

For the maximum value of $n$, substitute $D=0$,

$$\begin{gathered} D=0 \\ {b}^2-4ac=0 \\ {\left(2-n\right)}^2-4=0 \\ {n}^2-4n+4-4=0 \\ {n}^2=4n \\ n=4 \end{gathered}$$

Hence, the maximum value for the $n$ is $4$.