1. Given instantaneous voltage and current

$e = 100\sin 30t$ $i = 20\sin\left(30t-\frac{\pi}{4}\right)$

So,

• Peak voltage: $E_0 = 100$
• Peak current: $I_0 = 20$
• Phase difference: $\phi = \frac{\pi}{4}$

The current lags the voltage by $\frac{\pi}{4}$.

2. Find rms values

$V_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{100}{\sqrt{2}}$

$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{20}{\sqrt{2}}$

3. Average power consumed

For an a.c. circuit,

$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi$

Substitute the values:

$P_{\text{avg}} = \frac{100}{\sqrt{2}} \cdot \frac{20}{\sqrt{2}} \cdot \cos\frac{\pi}{4}$

$= \frac{2000}{2} \cdot \frac{1}{\sqrt{2}}$

$= \frac{1000}{\sqrt{2}}\,\text{W}$

4. Wattless current

Wattless current is the reactive component of current:

$I_w = I_{\text{rms}}\sin\phi$

So,

$I_w = \frac{20}{\sqrt{2}} \cdot \sin\frac{\pi}{4}$

$= \frac{20}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{20}{2} = 10\,\text{A}$

5. Final answer

The average power consumed and wattless current are

$\left(\frac{1000}{\sqrt{2}},\ 10\right)$

So the correct option is C.