Question

A network of four capacitors each off 12 µF capacitor and he’s connected to a 500 V supply as shown in figure determine the equivalent capacitance of the network add charge of each capacitor

Answer 1

Explanation:

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Answer 2

The equivalent capacitance is 16 μF.

The charge on C₁, C₂ , C₃ capacitor and C₄ is  $8\times10^{-3}\ C$ and $1.99\times10^{-3}\ C$

Explanation:

Given that,

The value of each capacitors = 12μF

Voltage = 500 V

According to figure,

C₁, C₂ and C₃ are connected in series

We need to calculate the equivalent capacitance

Using formula of series

$\dfrac{1}{C}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$

Put the value into the formula

$\dfrac{1}{C}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}$

$C=4\ \muF$

Now, C and C₄ is connecting in parallel

We need to calculate the equivalent capacitance

Using formula for parallel

$C'=C+C_{4}$

Put the value into the formula

$C'=12+4$

$C'=16\ \muF$

We need to calculate the charge C₁, C₂ and C₃

Using formula of charge

$Q=C_{eq}V$

Put the value in to the formula

$Q=16\times10^{-6}\times500$

$Q=8\times10^{-3}\ C$

We need to calculate the voltage across parallel capacitors

Using formula of voltage

$V'=\dfrac{Q}{C_{4}}$

Put the value into the formula

$V'=\dfrac{8\times10^{-3}}{12\times10^{-6}}$

$V'=666.6\ V$

We need to calculate the across voltage on C₄

$V''=V'-V$

$V''=666.6-500$

$V''=166.6\ V$

We need to calculate the charge on C₄

Using formula of charge

$Q'=CV''$

$Q'=12\times10^{-6}\times166.6$

$Q'=1.99\times10^{-3}\ C$

Hence, The equivalent capacitance is 16 μF.

The charge on C₁, C₂ , C₃ capacitor and C₄ is  $8\times10^{-3}\ C$ and $1.99\times10^{-3}\ C$

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Topic : charge

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