2.25      Obtain the equivalent capacitance of the network in Figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Given.

$C_1=100pF$

$C_2=200pF$

$C_3=200pF$

$C_4=100pF$

Now,

Lets first calculate the equivalent capacitance of $C_2\: and \:C_3$

$C_{23}=\frac{C_2C_3}{C_2+C_3}=\frac{200*200}{200+200}=100pF$

Now let's calculate the equivalent of $C_1\:and\:C_{23}$

$C_{1-23}=C_1+C_{23}=100+100=200pF$

Now let's calculate the equivalent of $C_{1-23} \: and \:C_4$

$C_{equivalent}=\frac{C_{1-23}*C_4}{C_{1-23}+C_4}=\frac{100*200}{100+200}=\frac{200}{3}pF$

Now,

The total charge on $C_4$ capacitors:

$Q_4=C_{equivalent}V=\frac{200}{3}*10^{-12}*300=2*10^{-8}C$

So,

$V_4=\frac{Q_4}{C_4}=\frac{2*10^{-8}}{100*10^{-12}}=200V$

The voltage across $C_1$ is given by

$V_{1}=V-V_{4}=300-200=100V$

The charge on $C_1$ is given by

$Q_1=C_1V_1=100*10^{-12}*100=10^{-8}C$

The potential difference across $C_2\:and\:C_3$ is

$V_2=V_3=50V$

Hence Charge on $C_2$

$Q_2=C_2V_2=200*10^{-12}*50=10^{-8}C$

And Charge on $C_3$:

$Q_3=C_3V_3=200*10^{-12}*50=10^{-8}C$

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