In an A.C. circuit, the instantaneous e.m.f. and current are given by, $$E = 100 \sin 30t$$, $$I = 20 \sin\left(30t - \frac{\pi}{4}\right)$$. In one cycle of A.C., the average power consumed by the circuit (in watt) and the watt-less current (in ampere) are, respectively:

We have an alternating e.m.f. (voltage) and current described by

$$E(t)=100\sin 30t,\qquad I(t)=20\sin\!\left(30t-\frac{\pi}{4}\right).$$

The amplitude (maximum value) of the voltage is therefore

$$E_{0}=100,$$

and the amplitude of the current is

$$I_{0}=20.$$

First we convert these amplitudes to their r.m.s. (root-mean-square) values. The standard relation is

$$\text{r.m.s. value}=\frac{\text{amplitude}}{\sqrt{2}}.$$

So for the voltage

$$V_{\text{rms}}=\frac{E_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}},$$

and for the current

$$I_{\text{rms}}=\frac{I_{0}}{\sqrt{2}}=\frac{20}{\sqrt{2}}.$$

Next we identify the phase difference between the voltage and the current. From the given expressions, the voltage is $$\sin 30t$$ while the current is $$\sin(30t-\pi/4).$$ The current therefore lags the voltage by

$$\phi=\frac{\pi}{4}\;(\text{that is, }45^{\circ}).$$

The average (or real) power consumed by an a.c. circuit is found with the formula

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

Substituting the numerical values we have just obtained:

$$P_{\text{avg}}=\left(\frac{100}{\sqrt{2}}\right)\!\left(\frac{20}{\sqrt{2}}\right)\cos\!\left(\frac{\pi}{4}\right).$$

Now evaluate step by step:

First multiply the numerators:

$$100\times20=2000.$$

Multiply the denominators:

$$\sqrt{2}\times\sqrt{2}=2.$$

Thus the product of the two r.m.s. values is

$$\frac{2000}{2}=1000.$$

Next, recall the trigonometric value

$$\cos\!\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}.$$

So the average power becomes

$$P_{\text{avg}}=1000\times\frac{1}{\sqrt{2}}=\frac{1000}{\sqrt{2}}\ \text{watt}.$$

This completes the calculation of the real (average) power.

Now we determine the watt-less (reactive) component of the current. The reactive current is the component that is out of phase by 90° with the voltage. Its magnitude is

$$I_{\text{watt-less}}=I_{\text{rms}}\sin\phi.$$

Again substituting the numbers:

$$I_{\text{watt-less}}=\left(\frac{20}{\sqrt{2}}\right)\sin\!\left(\frac{\pi}{4}\right).$$

We use the trigonometric value

$$\sin\!\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}.$$

Therefore

$$I_{\text{watt-less}}=\frac{20}{\sqrt{2}}\times\frac{1}{\sqrt{2}}=\frac{20}{2}=10\ \text{ampere}.$$

Thus, in one complete cycle of the given a.c. the circuit consumes an average power of $$\dfrac{1000}{\sqrt{2}}\,$$ watt and carries a watt-less current of $$10$$ ampere.

Hence, the correct answer is Option C.