What is the p.d of a capacitor with a capacitance of 3,300pF needs to store a 0.5mj of energy?

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What is the p.d of a capacitor with a capacitance of 3,300pF needs to store a 0.5mj of energy?

How to find the voltage and what is the formula for energy and also the power

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Answer 1 · 2017-05-05T00:00:56

$550.5 V$ $550.5V$ , rounded to two decimal places

Explanation:

Energy $U$ $U$ stored in a capacitor is given by

$U = \frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} Q V = \frac{1}{2} C V^{2}$ $U=1/2Q^2/C=1/2QV=1/2CV^2$
where $Q$ $Q$ is charge held, $C$ $C$ its capacitance and $V$ $V$ potential difference between the plates.

Using the expression of interest and inserting given values we get
$U = \frac{1}{2} C V^{2}$ $U=1/2CV^2$
$0.5 \times 10^{- 3} = \frac{1}{2} \times (3300 \times 10^{- 12}) V^{2}$ $0.5xx10^-3=1/2xx(3300xx10^-12)V^2$
$\Rightarrow V^{2} = 2 \times 0.5 \times \frac{10^{- 3}}{3300 \times 10^{- 12}}$ $=>V^2=2xx0.5xx10^-3/(3300xx10^-12)$
$\Rightarrow V^{2} = 2 \times 0.5 \times \frac{10^{- 3}}{3300 \times 10^{- 12}}$ $=>V^2=2xx0.5xx10^-3/(3300xx10^-12)$
$\Rightarrow V = 550.5 V$ $=>V=550.5V$ , rounded to two decimal places

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What is the p.d of a capacitor with a capacitance of 3,300pF needs to store a 0.5mj of energy?

How to find the voltage and what is the formula for energy and also the power

1 Answer

Explanation:

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