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

Kaspaueru2

Kaspaueru2

Answered question

2022-01-06

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.

Answer & Explanation

eskalopit

eskalopit

Beginner2022-01-07Added 31 answers

Ceq=C1+C2=2+4=6μF
U=(12)CV2=(12)(6e6)(300300)=0.27J
eninsala06

eninsala06

Beginner2022-01-08Added 37 answers

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(2.0×106F+4.0×106F)(300V)2=0.27J.
karton

karton

Expert2022-01-11Added 613 answers

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=(6×106)(3002)2E=0.27J
This is the energy stored in the circuit

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