Question

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Answer 1

Given:

A complex circuit consisting of capacitors have been provided.

To find:

The equivalent capacitance in between terminals A and B?

Calculation:

In this kind of complex circuits we first need to to designate potential drops along each capacitor and then try to simplify the circuit using the following tool :

• Capacitors which same potential drop will be considered in parallel combination.

Now, refer to the attached image to understand the designation of potential drops and the simplified circuit:

• Now, all the capacitors have capacitance of $4\mu F$.

Now, the simplified circuit can be calculated in series combination in each branch ( see 3rd diagram).

• Finally all the branches are in parallel.

$\sf \therefore \: C_{eq} = C_{1} + C_{2} + C_{3}$

$\sf \implies \: C_{eq} =4 + 2 + 2$

$\sf \implies \: C_{eq} =8 \: \mu F$

Hope It Helps.

Answer 2

Answer:

Given:

A complex circuit consisting of capacitors have been provided.

To find:

The equivalent capacitance in between terminals A and B?

Calculation:

In this kind of complex circuits we first need to to designate potential drops along each capacitor and then try to simplify the circuit using the following tool :

Capacitors which same potential drop will be considered in parallel combination.

Now, refer to the attached image to understand the designation of potential drops and the simplified circuit:

Now, all the capacitors have capacitance of 4\mu F4μF .

Now, the simplified circuit can be calculated in series combination in each branch ( see 3rd diagram).

Finally all the branches are in parallel.

\sf \therefore \: C_{eq} = C_{1} + C_{2} + C_{3}∴C

eq

=C

1 +C 2 +C 3

\sf \implies \: C_{eq} =4 + 2 + 2⟹C

eq

=4+2+2

\sf \implies \: C_{eq} =8 \: \mu F⟹C

eq

=8μF

Hope It Helps.