The variation of the instantaneous current (I | vedclass

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(B) From the given figure,the emf $(E)$ starts from zero at $\omega t = 0$ and reaches its maximum at $\omega t = \pi / 2$. Thus,$E = E_0 \sin(\omega

Vedclass · Accepted Answer

The voltage leads the current by $\pi / 2$.

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(B) From the given figure,the emf $(E)$ starts from zero at $\omega t = 0$ and reaches its maximum at $\omega t = \pi / 2$. Thus,$E = E_0 \sin(\omega t)$.The current $(I)$ starts from its minimum value at $\omega t = 0$ and reaches zero at $\omega t = \pi / 2$. This indicates that the current follows the equation $I = I_0 \sin(\omega t - \pi / 2)$.Comparing the phases,the phase of the voltage is $\omega t$ and the phase of the current is $(\omega t - \pi / 2)$.Therefore,the voltage leads the current by a phase angle of $\pi / 2$.

The variation of the instantaneous current $(I)$ and the instantaneous emf $(E)$ in a circuit is as shown in the figure. Which of the following statements is correct?

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