You are not allowed to take pictures of any question. Question 2 A 50-µF capacitor is charged to 400 V. Its stored energy is: 0.1 J 4J 0.4 J 4000 J 0.004 J No new data to save last checked at 0.04am
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In this case, the capacitance (C) is given as 50 μF (microfarads) and the voltage (V) is given as 400V. Plugging these values into the formula, we get: E = (1/2) * 50 * 10^-6 * (400)^2 Simplifying the equation, we have: E = (1/2) * 50 * 10^-6 * 160000 E = Show more…
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Sam S.
A $1.2-\mu \mathrm{F}$ capacitor is charged to $3.0 \mathrm{kV}$. Compute the energy stored in the capacitor. $$ \text { Energy }=\frac{1}{2} q V=\frac{1}{2} C V^{2}=\frac{1}{2}\left(1.2 \times 10^{-6} \mathrm{~F}\right)(3000 \mathrm{~V})^{2}=5.4 \mathrm{~J} $$
4. What is the energy stored in a 5 μF capacitors if the charge present in it is 30 μC? A. 30 μJ B. 90 μJ C. 150 μJ D. 540 μJ
Sri K.
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