In an $$a.c.$$  circuit the voltage applied is $$E = {E_0}\sin \omega t.$$    The resulting current in the circuit is $$I = {I_0}\sin \left( {\omega t - \frac{\pi }{2}} \right).$$     The power consumption in the circuit is given by

Solution :
KEY CONCEPT: We know that power consumed in $$a.c.$$  circuit is given by, $$P = {E_{rms}}.{I_{rms}}\cos \phi $$
Here, $$E = {E_0}\sin \omega t$$
$$I = {I_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$$
which implies that the phase difference, $$\phi = \frac{\pi }{2}$$
$$\therefore P = {E_{rms}}.{I_{rms}}\cos \frac{\pi }{2} = 0\,\,\,\,\,\left( {\because \cos \frac{\pi }{2} = 0} \right)$$