Question

The resistance of the series combination of two resistors is $\text{S}$. When they are joined in a parallel, the total resistance is $\text{P}$. If $$S=nP$$ the minimum possible value of $n$ is ?

• A. $4$
  
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $4$


Method $\left(I\right)$:

Let the resistors be $R$ and $xR$

Then $S=R+xR=(x+1)R$ and

$P=\frac{R\left(xR\right)}{R+xR}=\frac{xR}{x+1}$

$n=\frac{S}{P}=\frac{(x+1{)}^2}{x}=x+2+\frac{1}{x}$

For minimum $n$

$\frac{dn}{dx}=1-\frac{1}{x^2}=0$

$x=\pm 1$

We choose the +ve value of $x$

$\therefore x=1$

then the resistors are $R$ and $R$

$n_{min}=\frac{S}{P}=\frac{(1+1{)}^2}{1}=4$

Method $\left(II\right)$:

$n=\frac{\text{Resistance in series combination}}{\text{Resistance in parallel combination}}=\frac{S}{P}$

If the sum of two resistances is fixed then,

To obtain minimum value of $n$, resistance in parallel combination has to be maximum.

This can be achieved when the two resistances are equal.

In parallel combination of resistances :

Equivalent resistance $<$ lowest individual resistance


$\therefore R_{eq}=\frac{RR}{R+R}\Rightarrow R_{eq}=P=\frac{R}{2}$

$S=R+R=2R$

$\because S=nP$

$\therefore 2R=\frac{nR}{2}\Rightarrow n=4$

Hence, option (a) is the correct answer.