Question

In an $\text{AC}$ circuit, the instantaneous $\text{EMF}$ and current is given by $E=100\sin 30t,i=20\sin (30t-\frac{\pi }{4})$.
In one cycle of $\text{AC}$, the average power consumed by the circuit and the wattles current are, respectively

• A. $50\text{W},10\text{A}$
• B. $\frac{1000}{\sqrt{2}}\text{W},10\text{A}$
• C. $\frac{50}{\sqrt{2}}\text{W},0\text{A}$
• D. $50\text{W},0\text{A}$

Answer 1

The correct option is B $\frac{1000}{\sqrt{2}}\text{W},10\text{A}$

Given:

$E=100\sin 30t$, and

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

Average power consumed by the circuit is given by:

$P_{avg}=V_{rms}{I}_{rms}\cos \varphi$

Here,
$V_{rms}=\frac{100}{\sqrt{2}}\text{V},I_{rms}=\frac{20}{\sqrt{2}}\text{A}$

and $\varphi =[30t-(30t-\frac{\pi }{4})]=\frac{\pi }{4}$

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

$=\frac{2000}{2}\times \frac{1}{\sqrt{2}}=\frac{1000}{\sqrt{2}}\text{W}$

The wattles current is given by:

$I=I_{rms}\sin \varphi$

$I=\frac{20}{\sqrt{2}}\times \sin \left(\frac{\pi }{4}\right)$

$\therefore I=\frac{20}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=10\text{A}$

Hence, option $\left(\text{B}\right)$ is the correct answer.