Q17 of 26 Page 1

In the following arrangement of capacitors, the energy stored in the 6 μF capacitor is E. Find the value of the following :(i) Energy stored in 12 μF capacitor. (ii) Energy stored in 3 μF capacitor. (iii) Total energy drawn from the battery.


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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