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

${C}_{eq}={C}_{1}+{C}_{2}=2+4=6\mu F$
$U=\left(\frac{1}{2}\right)C{V}^{2}=\left(\frac{1}{2}\right)\cdot \left(6e-6\right)\cdot \left(300\cdot 300\right)=0.27J$

eninsala06

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

karton

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:
$E=\frac{C{V}^{2}}{2}\phantom{\rule{0ex}{0ex}}E=\frac{\left(6×{10}^{-6}\right)\left({300}^{2}\right)}{2}\phantom{\rule{0ex}{0ex}}E=0.27J$
This is the energy stored in the circuit

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