Question

If a capacitor between A and B is $12\mu F$ the capacity C is?
A. $5\mu F$
B. $3\mu F$
C. $4\mu F$
D. $8\mu F$

Answer 1

we will solve this question by finding the equation of total capacitance of the circuit. Both $6\mu F$ capacitors are in series. Both $8\mu F$ capacitors are also in series with each other. And the unknown capacity C is in parallel with them. The equivalent capacitance has been given to us, thus we will equate the given capacitance to the equivalent capacitance found by us after applying the formulas.

Formula used:
Equivalent capacitance of 2 capacitors in series is given by:
$\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Equivalent capacitance of 2 capacitors in parallel is given by:
${C_{eq}} = {C_1} + {C_2}$

Complete step by step answer:
Look at the diagram below to get a better idea:

Let us start solving the question by finding equivalent capacitance of $6\mu F$ in series:
$\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{6} + \dfrac{1}{6}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{2}{6}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{3}$
$\Rightarrow {C_{eq}} = 3\mu F$

Now we will find equivalent capacitance of $8\mu F$ capacitors:
$\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{8} + \dfrac{1}{8}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{2}{8}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{4}$
$\Rightarrow {C_{eq}} = 4\mu F$.

Now these 3 capacitors ${C_{eq}} = 4\mu F$, ${C_{eq}} = 3\mu F$ and $C\mu F$ are in parallel connection. Hence now we will apply the formula for equivalent capacitance of capacitors in parallel.
${C_{eq}} = {C_1} + {C_2} + {C_3}$
$\Rightarrow {C_{eq}} = C + 3 + 4$
$\Rightarrow {C_{eq}} = C + 7$
It has been mentioned that equivalent capacitance is $12\mu F$.
Hence we will substitute ${C_{eq}} = 12$
$12 = C + 7$
$\Rightarrow C = 12 - 7\mu F$
$\therefore C = 5\mu F$.
Hence the capacity of capacitor C is $5\mu F$.

Hence the correct answer is option A.

Note: Always remember that the formula for equivalent capacitance in series and parallel combination of capacitors is opposite to the formula of resistors. We said that the $6\mu F$ and $8\mu F$ capacitors are in series because the potential across the ends of capacitors is different. And later $8\mu F$, $6\mu F$ and C are said to be parallel because the potential difference across their ends is the same.