Question 282: AC fundamentals

Question

In an $L-R$ circuit, the inductive reactance is equal to $\sqrt{3}$ times the resistance ' $R$ ' of the circuit. An e.1n.f. $\mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t})$ is applied to the circuit. The power consumed in the circuit is

Accepted Answer

Given:
Inductive reactance, $X_L = \sqrt{3} R$
Applied emf, $E = E_0 \sin(\omega t)$
LetÃ¢ÂÂs proceed step by step:
1. Impedance of the Circuit
The total impedance ($Z$) of an $L-R$ circuit is:
$$ Z = \sqrt{R^2 + X_L^2} $$
Given $X_L = \sqrt{3} R$, so:
$$ Z = \sqrt{R^2 + (\sqrt{3}R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R $$
2. Expression for rms values
The root mean square (rms) value of the applied emf:
$$ E_{	ext{rms}} = \frac{E_0}{\sqrt{2}} $$
The rms current in the circuit:
$$ I_{	ext{rms}} = \frac{E_{	ext{rms}}}{Z} = \frac{E_0/\sqrt{2}}{2R} = \frac{E_0}{2R\sqrt{2}} $$
3. Power Consumed
The average power consumed is:
$$ P = E_{	ext{rms}} \cdot I_{	ext{rms}} \cdot \cos\phi $$
Where $\cos\phi$ is the power factor, $\phi$ is the phase angle:
$$ 	an\phi = \frac{X_L}{R} = \frac{\sqrt{3}R}{R} = \sqrt{3} $$
So,
$$ \phi = 60^\circ $$
Thus,
$$ \cos\phi = \cos 60^\circ = \frac{1}{2} $$
4. Substitute and Simplify
$$ P = E_{	ext{rms}} \cdot I_{	ext{rms}} \cdot \frac{1}{2} $$
Substitute the values:
$$ P = \left(\frac{E_0}{\sqrt{2}}ight) \left(\frac{E_0}{2R\sqrt{2}}ight) \frac{1}{2} $$
Multiply the numerators and denominators:
$$ = \frac{E_0 \cdot E_0}{\sqrt{2} \cdot 2R \cdot \sqrt{2}} \cdot \frac{1}{2} $$
$$ = \frac{E_0^2}{2 \cdot 2R} \cdot \frac{1}{2} $$
$$ = \frac{E_0^2}{4R} \cdot \frac{1}{2} $$
$$ = \frac{E_0^2}{8R} $$
Final Answer:
The power consumed in the circuit is
$ \boxed{\frac{E_0^2}{8R}} $
So, the correct option is
Option C
.

Answer

$\frac{\mathrm{E}_0^2}{4 \mathrm{R}}$

Answer

$\frac{E_0^2}{6 R}$

Answer

$\frac{\mathrm{E}_0^2}{8 \mathrm{R}}$

Answer

$\frac{E_0^2}{12 R}$

AC fundamentals

In an L-R circuit, the inductive reactance is equal to \sqrt{3} times the resistance ' R ' of the circuit. An e.1n.f. \mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t}) is applied to the circuit. The power consumed in the circuit is

Choose the correct answer: