Question

The equivalent capacitance between A and B for the combination of capacitors shown in figure, where all capacitances are in microfarad is

Answer 1

Hint: Express the mathematical formula to find the equivalent capacitance of more than one capacitor connected in series and parallel to each other. Simplify the circuit starting from the smallest loop and at last we will find our answer.

Complete step by step answer:
We can find the equivalent capacitance of more than one capacitor depending on the connection of the capacitors.
If two capacitors ${{C}_{1}}\And {{C}_{2}}$ are connected in series than the equivalent capacitance C can be found out as,

$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$

If two capacitors ${{C}_{1}}\And {{C}_{2}}$ are connected in parallel than the equivalent capacitance will be,

$C={{C}_{1}}+{{C}_{2}}$

Now, looking at the given circuit,

In the top part of the circuit we have a $1\mu F$ and one $3\mu F$ capacitor in parallel.
Let the equivalent capacitance will be ${{C}_{1}}$
So,

$\begin{aligned} & {{C}_{1}}=1\mu F+3\mu F \\\ & {{C}_{1}}=4\mu F \\\ \end{aligned}$

Again, at the bottom of the circuit we have one $6\mu F$ and one $2\mu F$ capacitor connected in parallel.
Let the equivalent capacitance is ${{C}_{2}}$
So,

$\begin{aligned} & {{C}_{2}}=6\mu F+2\mu F \\\ & {{C}_{2}}=8\mu F \\\ \end{aligned}$

So, the equivalent circuit becomes,

Now, at the top of the circuit we have one $4\mu F$ and one $4\mu F$ capacitor connected in series.

Let the equivalent capacitance is ${{C}_{3}}$
So,

$\begin{aligned} & \dfrac{1}{{{C}_{3}}}=\dfrac{1}{4\mu F}+\dfrac{1}{4\mu F} \\\ & \dfrac{1}{{{C}_{3}}}=\dfrac{2}{4\mu F} \\\ & {{C}_{3}}=2\mu F \\\ \end{aligned}$

Again, at the bottom of the circuit we have one $8\mu F$ and one $8\mu F$ capacitor connected in series.

Let the equivalent capacitance is ${{C}_{4}}$.
So,

$\begin{aligned} & \dfrac{1}{{{C}_{4}}}=\dfrac{1}{8\mu F}+\dfrac{1}{8\mu F} \\\ & \dfrac{1}{{{C}_{4}}}=\dfrac{2}{8\mu F} \\\ & {{C}_{4}}=4\mu F \\\ \end{aligned}$

So, the equivalent circuit becomes

Now we have one $4\mu F$ and one $2\mu F$ capacitor in parallel. Let the equivalent capacitance is C.
So,

$\begin{aligned} & C=4\mu F+2\mu F \\\ & C=6\mu F \\\ \end{aligned}$

So, the equivalent capacitance of the circuit is $6\mu F$.

Note: Capacitor is a device which stores electrical energy in presence of an electric field. Capacitors in series are the same as resistance in parallel while resistors in series are the same as capacitors in parallel.