1. Given expressions

   The applied voltage is $E = E_0\sin \omega t$

   The current is $I = I_0\sin\left(\omega t - \frac{\pi}{2}\right)$

2. Identify the phase difference

   Comparing voltage and current:

   • Voltage phase $= \omega t$
   • Current phase $= \omega t - \frac{\pi}{2}$

   So the current lags the voltage by $\phi = \frac{\pi}{2}$

3. Use average power formula in AC circuit

   The average power consumed in an AC circuit is $P_{\text{avg}} = E_{\text{rms}} I_{\text{rms}} \cos\phi$

   where $E_{\text{rms}} = \frac{E_0}{\sqrt{2}}, \qquad I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

   Therefore, $P_{\text{avg}} = \frac{E_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cdot \cos\frac{\pi}{2}$

   Since $\cos\frac{\pi}{2} = 0$

   we get $P_{\text{avg}} = 0$

4. Alternative verification using instantaneous power

   Instantaneous power is $p = EI = E_0\sin\omega t \cdot I_0\sin\left(\omega t - \frac{\pi}{2}\right)$

   Using $\sin\left(\omega t - \frac{\pi}{2}\right) = -\cos\omega t$

   $p = -E_0I_0\sin\omega t\cos\omega t$

   Over one complete cycle, the average of $\sin\omega t\cos\omega t$ is zero, so average power is again $P=0$

5. Check options

   • A: $P=\sqrt{2}E_0I_0$ ❌
   • B: $P=\dfrac{E_0I_0}{\sqrt{2}}$ ❌
   • C: $P=\text{zero}$ ✅
   • D: $P=\dfrac{E_0I_0}{2}$ ❌

Hence, the correct option is C.