Q. 17

# In the following

Given:

Capacitor, C1=6 μF

Capacitor, C2=12 μF

Capacitor, C3=3 μF

Calculation:

The energy stored in the capacitor is given by,

E=CV2/2

Where,

E = energy

C= capacitance

V=Voltage

For capacitor C1:

The energy stored in the capacitor is given by,

E= (6 μF) V2/2

I. For capacitor C2:

The energy stored in the capacitor is given by,

E2= (12 μF) V2/2 = 2E

II. For capacitor C3:

The charge on capacitor C3 will be equal to the charge flowing through the parallel combination of C1 and C2,

Q=C12+6V=C3V1

V1= 6 V

E3=(3 μF)(6V)2/2

E3= 18E

III. Total energy is given by,

E=E+2E+18E=21E

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