MHT CET · Physics · Alternating Current
In an a.c. circuit with pure capacitance 'C' and a.c. source \(\mathrm{E}=\mathrm{E}_0 \sin \omega \mathrm{t}\), the equation of instantaneous current is given by
- A \(I=E_0 \omega c \sin (\omega t)\)
- B \(I=E_0 \omega c \sin \left(\omega t+\frac{\pi}{2}\right)\)
- C \(I=\frac{E_0}{\omega c} \sin (\omega t)\)
- D \(I=\frac{E_0}{\omega c} \sin \left(\omega t+\frac{\pi}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(I=E_0 \omega c \sin \left(\omega t+\frac{\pi}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(I_0 = E_0 \omega C\) \(I = I_0 \sin \left(\omega t+\frac{\pi}{2}\right)\)
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