Question

The figure shows a network of capacitors where the numbers indicate capacitances in micro Farad. The value of capacitance C such that the equivalent capacitance between point A and B is to be $1\mu F$ is:

Answer 1

To solve the given question we must know when the capacitors are series and when they are parallel. Then we must also know the concept of effective capacitance of ‘n’ capacitors connected in series and effective capacitance of ‘n’ capacitors connected in parallel.

Formula used:
${{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+......+{{C}_{n}}$
$\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}+....+\dfrac{1}{{{C}_{n}}}$

Complete step by step answer:
Let us first check which of the capacitors are in parallel and find their effective capacitance first. When two more capacitors are across the potential difference or they have the same positive and negative terminals, the capacitors are said to be in parallel connection.
When the n capacitors of capacitances ${{C}_{1}},{{C}_{2}},{{C}_{3}},......{{C}_{n}}$ are in parallel connection, the effective capacitance of the combination is given as ${{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+......+{{C}_{n}}$.
In the given figure, we can see that the two capacitors of capacitance $2\mu F$ in parallel connection.
Therefore, the effective capacitance of the two is ${{C}_{eq}}=2\mu +2\mu =4\mu F$.
When the n capacitors of capacitances ${{C}_{1}},{{C}_{2}},{{C}_{3}},......{{C}_{n}}$ are in series connection, the effective capacitance of the combination is given as $\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}+....+\dfrac{1}{{{C}_{n}}}$.
In the given figure, we can see that the capacitors of capacitance $6\mu F$ and $12\mu F$ in series connection.
Then we can write that $\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{6\mu }+\dfrac{1}{12\mu }=\dfrac{1}{4\mu }$
Therefore, the effective capacitance of the two is ${{C}_{eq}}=4\mu F$.
With this the circuit can be simplified as shown below.

Now, we can see that the two $4\mu F$ capacitors are in parallel. Therefore, the effective capacitance of these two is ${{C}_{eq}}=4\mu +4\mu =8\mu F$.
And the capacitors of capacitances $8\mu F$ and $4\mu F$ in series connection. Therefore, the effective capacitance of these two is $\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{8\mu }+\dfrac{1}{4\mu }=\dfrac{3}{8\mu }$.
$\Rightarrow {{C}_{eq}}=\dfrac{8\mu }{3}F$
The circuit can be simplified further as shown.

Now, the capacitors of capacitances $1\mu F$ and $8\mu F$ in series connection. Therefore, the effective capacitance of these two is $\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{1\mu }+\dfrac{1}{8\mu }=\dfrac{9}{8\mu }$.
$\Rightarrow {{C}_{eq}}=\dfrac{8\mu }{9}F$.

Now, the capacitors of capacitances $\dfrac{8}{3}\mu F$ and $\dfrac{8}{9}\mu F$ in parallel connection. Therefore, the effective capacitance of these two is ${{C}_{eq}}=\dfrac{8}{3}\mu +\dfrac{8}{9}\mu =\dfrac{32}{9}\mu F$.
Therefore, the circuit simplifies to:

Therefore, the effective capacitance of the whole circuit is $\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{C}+\dfrac{9}{32\mu }$.
But it is given that the effective capacitance of the whole circuit is $1\mu F$.
$\Rightarrow \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{C}+\dfrac{9}{32\mu }=\dfrac{1}{1\mu }$
$\Rightarrow \dfrac{1}{C}=\dfrac{1}{1\mu }-\dfrac{9}{32\mu }=\dfrac{23}{32\mu }$
$\Rightarrow C=\dfrac{32\mu }{23}F$
Therefore, the required value of capacitance C is $\dfrac{32}{23}\mu F$

Note:
The formulae for the effective capacitance of n capacitors and effective resistance of n resistors can be misunderstood.
The formula for effective resistance of n resistances in series is similar to the formula for effective capacitance of n capacitors in parallel.
The formula for effective resistance of n resistances in parallel is similar to the formula for effective capacitance of n capacitors in series.