Question

Find equivalent capacitance between points A and B for the given network. (Given: C1=C2=C3= 1uF & C4 = 2uF )

Answer 1

The equivalent capacitance between points A and B for the given network is $\dfrac{4}{5}\ \mu F$

Explanation:

Given that,

Capacitance, $C_1=C_2=C_3=1\ \mu F$

Capacitance, $C_4=2\ \mu F$

We need to find the equivalent capacitance between points A and B for the given network. It is clear that, capacitors $C_2\ and\ C_4$ are in parallel. The equivalent is given by :

$C'=C_2+C_4$

$C'=1+2=3\ \mu F$

Now $C_3\ and\ C'$ are again in parallel. So,

$C''=C'+C_3$

$C''=1+3=4\ \mu F$

Now, $C''\ and\ C_1$ are in series. So,

$\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C''}$

$\dfrac{1}{C_e}=\dfrac{1}{1}+\dfrac{1}{4}$

$C_e=\dfrac{4}{5}\ \mu F$

So, the equivalent capacitance between points A and B for the given network is $\dfrac{4}{5}\ \mu F$

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