A 2.0μF capacitor and a 4.0μF capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.

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eskalopit · Accepted Answer

Ceq=C1+C2=2+4=6μF  U=(12)CV2=(12)⋅(6e−6)⋅(300⋅300)=0.27J

eninsala06 · Accepted Answer

The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference V across the capacitors is the same and the total energy is :  U=12(C1+C2)V2=12(2.0×10−6F+4.0×10−6F)(300V)2=0.27J.

karton · Accepted Answer

Step 1 Solution: Given that C1=2micro FC2=4micro FV=300V Step 2 As given the two capacitors are connected in parallel to each other, hence the total capacitance of the circuit is the sum of both capacitance, which can be given as below: the total capacitance of the circuit is given as C=C1+C2C=(2+4)micro FC=6micro F Now the energy stored is given by: E=CV22E=(6×10−6)(3002)2E=0.27J This is the energy stored in the circuit

A 2.0 \mu F capacitor and a 4.0 \mu F

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