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

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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 $\pi / 2$.

Currents	$r.m.s.$ values
$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2\sqrt{2}}$

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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