$H = \frac{V^2}{R} e^{-2t/RC} dt$
From 0 to ∞.
I put $k = e^{-2t/RC}$
I tried taking log on both sides but got, something like $log k - t/2k = -t/2RC$
How can I solve it
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1\begin{align} \int_0^M \frac{V^2}{R} e^{-2t/RC} dt&=\left.\frac{V^2}{R}\left(-\frac{RC}2e^{-\frac{2}{RC}t}\right)\right|_0^M \end{align} – Ángel Mario Gallegos Apr 20 '16 at 16:51
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\int_{a}^{b} = $\int_a^b$ – Apr 20 '16 at 16:52
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@ÁngelMarioGallegos Thank you very much. – Anubhav Goel Apr 20 '16 at 16:54
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1"Taking $\log$"? Definitely a bad idea. How about $\int_a^b e^t,dt=e^b-e^a$ paired with a correct use of change of variables? – Apr 20 '16 at 16:54
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I forgot all maths after 1 month of my exam . So , silly of me. Thank you all. – Anubhav Goel Apr 20 '16 at 17:00
1 Answers
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\begin{align*} \int_0^M \frac{V^2}{R} e^{-2t/RC} dt&=\left.\frac{V^2}{R}\left(-\frac{RC}2e^{-\frac{2}{RC}t}\right)\right|_0^M\\[3pt] &=\frac{V^2C}{2}\left(1-e^{-\frac{2}{RC}t}\right) \end{align*} So \begin{align*} \int_0^{\infty} \frac{V^2}{R} e^{-2t/RC} dt&=\frac{V^2C}{2}\lim_{M\to\infty}\left(1-e^{-\frac{2}{RC}M}\right)\\[3pt] &=\frac{1}{2}CV^2 \end{align*}
Ángel Mario Gallegos
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