Kaspaueru2

2022-01-06

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

eskalopit

Beginner2022-01-07Added 31 answers

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 :

$U=\frac{1}{2}({C}_{1}+{C}_{2}){V}_{2}=\frac{1}{2}(2.0\times {10}^{-6}F+4.0\times {10}^{-6}F){\left(300V\right)}^{2}=0.27J$ .

karton

Expert2022-01-11Added 613 answers

Step 1

Solution:

Given that

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

Now the energy stored is given by:

This is the energy stored in the circuit

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