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]$
$\Rightarrow R = \frac{R_{1}C-R_{1}^{2}}{C} = R_{1}-\frac{R^{2}_{1}}{C}$
$\Rightarrow \frac{dR}{dR_{1}} = 1-\frac{2R_{1}}{C}$
and $\frac{d^{2}R}{dR_{1}^{2}} = -\frac{2}{C}$
For maximum or minimum, we must have
$\frac{dR}{dR_{1}} = 0$
$\Rightarrow 1-\frac{2R_{1}}{C} = 0$
$R_{1} = \frac{C}{2}
\frac{d^{2}R}{dR_{1}^{2}} = -\frac{2}{C} < 0$ 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 $R_{1} = R_{2} = \frac{C}{2}$