Question 274: AC fundamentals

Question

In an a.c. circuit with pure capacitance ' C ' and a.c. source $E=E_0 \sin \omega t$, the equation of instantaneous current is given by

Accepted Answer

Step 1: Voltage across capacitor
The applied AC source is
$ E(t) = E_0 \sin(\omega t). $
For a capacitor,
$ i(t) = C \frac{dV}{dt}. $
Thus,
$ i(t) = C \frac{d}{dt} \big[ E_0 \sin(\omega t) \big]. $
Step 2: Differentiate
$ \frac{d}{dt} \sin(\omega t) = \omega \cos(\omega t). $
So,
$ i(t) = C \cdot E_0 \cdot \omega \cos(\omega t). $
Step 3: Express in sine form
Recall:
$ \cos(\omega t) = \sin\left(\omega t + 	frac{\pi}{2}ight). $
Hence,
$ i(t) = E_0 \omega C \sin\!\left(\omega t + 	frac{\pi}{2}ight). $
Step 4: Compare with options
This matches
Option B
:
$ I = E_0 \omega C \sin\left(\omega t + 	frac{\pi}{2}ight). $
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Answer: Option B

Answer

$\mathrm{I}=\mathrm{E}_0 \omega \mathrm{c} \sin (\omega \mathrm{t})$

Answer

$I=E_0 \omega c \sin \left(\omega t+\frac{\pi}{2}\right)$

Answer

$\mathrm{I}=\frac{\mathrm{E}_0}{\omega \mathrm{c}} \sin (\omega \mathrm{t})$

Answer

$\mathrm{I}=\frac{\mathrm{E}_0}{\omega \mathrm{c}} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)$

AC fundamentals

In an a.c. circuit with pure capacitance ' C ' and a.c. source E=E_0 \sin \omega t, the equation of instantaneous current is given by

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