Advertisements
Advertisements
प्रश्न
Which expression gives the capacitor current in terms of peak current \[i_m\]?
पर्याय
\[i=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\]
\[i=i_m\sin\omega t\]
\[i=i_m\sin 2\omega t\]
\[i=i_m\sin\left(\omega t+\frac{\pi}{2}\right)\]
MCQ
Advertisements
उत्तर
The current is \[i=\omega Cv_m\cos\omega t\]. Rewriting cosine as a shifted sine gives \[i=i_m\sin\left(\omega t+\frac{\pi}{2}\right)\].
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
