Question

The combined resistance $R$ of two resistors $R_1$ and $R_2(R_1,R_2>0)$ is given by $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
If $R_1+R_2-C$ (a constant), then maximum resistance $R$ is obtained if

• A. $R_1>R_2$
• B. $R_1<R_2$
• C. $R_1=R_2$
• D. None of these

Answer 1

Answer: (C)

We have,
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
and $R_1+R_2=C$
$\Rightarrow \frac{1}{R}=\frac{R_1+R_2}{R_1{R}_2}=\frac{C}{R_1{R}_2}$
$=\frac{C}{R_1\left(C-R_1\right)}\left[\because R_2=C-R_1\right]$
$\Rightarrow R=\frac{R_{1}C-R_1^2}{C}=R_1-\frac{R_1^2}{C}$
$\Rightarrow \frac{dR}{d{R}_1}=1-\frac{2R_1}{C}$
and $\frac{d^{2}R}{d{R}_1^2}=-\frac{2}{C}$
For maximum or minimum, we must have
$\frac{dR}{d{R}_1}=0$
$\Rightarrow 1-\frac{2R_1}{C}=0$
$R_1=\frac{C}{2}<br/>\frac{d^{2}R}{d{R}_1^2}=-\frac{2}{C}<0$ for all values of $R_1$
Thus, $R$ is maximum when $R_1=\frac{C}{2}$
Putting $R_1=\frac{C}{2}$ in $R_1+R_2=C$,
we get $R_2$, $=C-\frac{C}{2}=\frac{C}{2}$
Hence, $R$ is maximum when $R_1=R_2=\frac{C}{2}$