In this circuit C1 8 mu F C2 4 mu F C3 4 mu F C4 6 mu F

In this circuit, C_1 = 8 mu F, C_2 = 4 mu F, C_3 = 4 mu F, C_4 = 6 mu F and V = 100 V. Find the equivalent capacitance. Find the charge on each capacitor and the voltage across each capacitor. C_2 suddenly shorts out so that it is effectively a conducting wire. What is the new equivalent capacitance for the circuit? Explain.

Solution

a)

C1 and C2 are in parallel

C₁₂ = 8+4 =12 uF

C12 ,C3 and C4 are in series ,so equivalent capacitance

1/C_eq =1/12 + 1/4 + 1/6

C_eq =2 uF

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b)

Total Chage

Q=C_eqV =100*2 uF

Q=200 uC

Charge on each capacitor

Q₁ =Q*(C₁/C₁+C₂) =200*(8/8+4) =133.33 uC

Q₂=Q*(C₂/C₁+C₂) =200*(4/8+4)=66.67 uC

Q₃=Q=200 uC

Q₄=Q=200 uC

Voltage across each capacitor

V₁=Q₁/C₁ =133.33/8 =16.67 Volts

V₂=Q₂/C₂ =66.67/4 =16.67 Volts

V₃=Q₃/C₃ =200/4 =50 Volts

V₄=Q₄/C₄ =200/6 =33.33 Volts

c)

Since C₂ is shorted ,Charge will only flow through Shorted C2 ,C3 and C4 .So C1 will not be considered in equivalent capacitance .So equivalent capacitance is

1/C_eq =1/C₃ + 1/C₄ =1/4+ 1/6

C_eq=2.4 uF