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julia-pushkina [17]
3 years ago
7

A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ

y is then stored in this capacitor combination?
Physics
1 answer:
Alborosie3 years ago
4 0

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, C_1=15\ \mu F

Capacitance, C_2=25\ \mu F

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F

The energy stored in the capacitor is given by :

E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

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In order to make any headway with this one, it might help
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I went and looked it up on line, you're welcome.

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