Question 38: Electrostats & Capacitance

Question

By connecting two given capacitors, a technician was able to make two new capacitors having the effective capacitance $12.5 \mu \mathbf{F}$ and $2 \mu \mathbf{F}$. What would be the capacitance of the given capacitors?

Accepted Answer

Let the two unknown capacitors be $C_1$ and $C_2$. When theyÃ¢ÂÂre connected in parallel, their capacitance is
$$ C_{m p} = C_1 + C_2 = 12.5\;\mu{m F}, $$
and in series itÃ¢ÂÂs
$$ C_{m s} = \frac{C_1 C_2}{C_1 + C_2} = 2\;\mu{m F}. $$
From the series formula we get
$$ C_1 C_2 = C_{m s}(C_1 + C_2) = 2 	imes 12.5 = 25\;(\mu{m F})^2. $$
So $C_1$ and $C_2$ are roots of
$$ t^2 - (C_1 + C_2)t + C_1 C_2 = t^2 - 12.5\,t + 25 = 0. $$
Its solutions are
$$ t = \frac{12.5 \pm \sqrt{12.5^2 - 4\cdot25}}{2}     = \frac{12.5 \pm 7.5}{2}, $$
i.e.\ $t=10$ or $t=2.5$.
Therefore the two capacitors are
10 ÃÂ¼F and 2.5 ÃÂ¼F, which is Option B.

Answer

$8.5 \mu \mathrm{~F}$ and $4 \mu \mathrm{~F}$

Answer

$10 \mu \mathrm{~F}$ and $2.5 \mu \mathrm{~F}$

Answer

$6.5 \mu \mathrm{~F}$ and $6 \mu \mathrm{~F}$

Answer

$10.5 \mu \mathrm{~F}$ and $2 \mu \mathrm{~F}$

Electrostats & Capacitance

By connecting two given capacitors, a technician was able to make two new capacitors having the effective capacitance 12.5 \mu \mathbf{F} and 2 \mu \mathbf{F}. What would be the capacitance of the given capacitors?

Choose the correct answer: