Question

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

• A. $\left(\frac{50}{\sqrt{2}},0\right)$
• B. $50,0$
• C. $50,10$
• D. $\left(\frac{1000}{\sqrt{2}},10\right)$

Answer 1

The correct option is D

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

Explanation for correct option:

Step 1: Given data

The instantaneous e.m.f. is, $\mathrm{e}=100\sin 30\mathrm{t}$

the instantaneous current is, $\mathrm{i}=20\sin \left(30\mathrm{t}-\left(\frac{\mathrm{\pi }}{4}\right)\right)$

We have to find the average power consumed and wattless current in one cycle.

Step 2: Formula to be used

In an AC circuit, the average power is determined by using root mean square current $\mathrm{rms}$ values of emf and current.

When the average power consumed by an AC circuit is equal to zero, wattless current is the amount of current that is flowing. The $\mathrm{rms}$ value of the current can be used to calculate it.

The average power consumed by an $\mathrm{A}.\mathrm{C}.$ circuit is,

${\mathrm{P}}_{\mathrm{avg}}={\mathrm{e}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}\mathrm{cos\varphi }$

Here, ${\mathrm{e}}_{\mathrm{rms}}$ is the average value of e.m.f. in the $\mathrm{A}.\mathrm{C}.$ circuit, ${\mathrm{I}}_{\mathrm{rms}}$ is the average value in the $\mathrm{A}.\mathrm{C}.$ and $\mathrm{\varphi }$ is the phase constant.

Wattless current is defined as the current in an $\mathrm{A}.\mathrm{C}.$ circuit when the average power consumed by the $\mathrm{A}.\mathrm{C}.$ circuit is equal to zero.

The wattles current is,

$\mathrm{I}={\mathrm{I}}_{\mathrm{rms}}\mathrm{sin\varphi }$

Here, $\mathrm{I}$ is the wattles current, ${\mathrm{I}}_{\mathrm{rms}}$ is the average value of current in an $\mathrm{A}.\mathrm{C}.$ circuit, and $\mathrm{\varphi }$ is the phase constant.

Step 3: Set the value of ${\mathrm{I}}_{\mathrm{rms}}$ and ${\mathrm{e}}_{\mathrm{rms}}$,

The values of instantaneous emf and current is given as,

$\mathrm{e}=100\sin 30\mathrm{t}$ ------- $\left(1\right)$

$\mathrm{i}=20\sin \left(30\mathrm{t}-\left(\frac{\mathrm{\pi }}{4}\right)\right)$ ------- $\left(2\right)$

So, these above two equations are in the form of wave equations due to dealing with alternating current and voltage.
From these wave equations, the maximum values of emf and current in the given $\mathrm{A}.\mathrm{C}.$ circuit is $100$ and $20$, respectively.

Let,

${\mathrm{e}}_{\max }=100\mathrm{V}$

${\mathrm{I}}_{\max }=20\mathrm{A}$

Here, ${\mathrm{e}}_{\max }$ is the maximum value of emf in the given $\mathrm{A}.\mathrm{C}.$ circuit, and ${\mathrm{I}}_{\max }$ is the maximum value of current
To find out the value of the average power consumed by the $\mathrm{A}.\mathrm{C}.$ circuit, we have to take $\mathrm{rms}$ values of emf and current.
We know that,

${\mathrm{e}}_{\mathrm{rms}}=\frac{{\mathrm{e}}_{\max }}{\sqrt{2}}$

and

${\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{I}}_{\max }}{\sqrt{2}}$
Substitute the values of ${\mathrm{e}}_{\max }$ and ${\mathrm{I}}_{\max }$ in the above set of equations, we get,

${\mathrm{e}}_{\mathrm{rms}}=\frac{{\mathrm{e}}_{\max }}{\sqrt{2}}$

$=\frac{100}{\sqrt{2}}$

and

${\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{I}}_{\max }}{\sqrt{2}}$

$=\frac{20}{\sqrt{2}}$ ------- $\left(3\right)$

Step 4. Calculation for average power consumed
Now, the average power consumed by an $\mathrm{A}.\mathrm{C}.$ circuit is,

${\mathrm{P}}_{\mathrm{avg}}={\mathrm{e}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}\mathrm{cos\varphi }$ ------- $\left(4\right)$
So, the phase constant is,

$\mathrm{\varphi }=\frac{\mathrm{\pi }}{4}$
Substitute this value as well as the values from the set of equations in $\left(3\right)$, in equation $\left(4\right)$, we get

${\mathrm{P}}_{\mathrm{avg}}={\mathrm{e}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}\mathrm{cos\varphi }$

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

$=1000\cos \frac{\mathrm{\pi }}{4}$

$=\frac{1000}{\sqrt{2}}$
Therefore, the average power consumed by the $\mathrm{A}.\mathrm{C}.$ circuit is,

${\mathrm{P}}_{\mathrm{avg}}=\frac{1000}{\sqrt{2}}$ ------- $\left(5\right)$

Step 5: Calculation for wattless current

We know that,

$\mathrm{I}={\mathrm{I}}_{\mathrm{rms}}\mathrm{sin\varphi }$

Substitute the values of ${\mathrm{I}}_{\mathrm{rms}}$ and $\mathrm{\varphi }$ in the above equation, we get

$\mathrm{I}={\mathrm{I}}_{\mathrm{rms}}\mathrm{sin\varphi }$

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

$=\frac{20}{\sqrt{2}\times \sqrt{2}}$

$=10$ -------$\left(6\right)$
Therefore, the amount of wattless current in the given AC circuit is $\mathrm{I}=10$
From equation $\left(5\right)$ and equation $\left(6\right)$, the average power consumed and the wattless current in one cycle of the given AC circuit are $\frac{1000}{\sqrt{2}}$ and $10$ respectively.

Hence, option (D) is the correct answer.