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

                

Answers (1)

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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