Question

A voltage v(t) = 100sin314t is applied to series circuit consisting of 100 ohm resistance,
0.0318H inductance and a capacitor of 63.6 micro Fared. Calculate
i) Expression for i(t)
ii) Phase angle between voltage and current
iii) Power factor
iv) active or real power consumed​

Answer 1

7888.633

578(43;.688)



thanks

Answer 2

Answer:

i) Expression for i(t) is$\\i(t) &=2.421 \sin \left(314 t+76^{\circ}\right) \mathrm{A}\end{aligned}$

ii) Phase angle between voltage and current $\theta=76^{0}$

iii) Power factor is $Y_{f}=0.242$

iv) active or real power consumed​  $2000 W$

Explanation:

We are authorized to answer three subparts at a time, since you have not mentioned which part you are looking for, so we are answering the first three subparts, please repost your question separately for the remaining subpart.

Given,

Series RLC circuit

$\begin{aligned}&e(t)=100 \sin 314 t \\&R=100 \Omega \\&L=0.0318 \mathrm{H} \\&C=63.6 \mu \mathrm{F}\end{aligned}$

The voltage applied is,

$e(t)=100 \sin 314 t$

The inductive reactance is, $x_{L}=g \omega L$

$\begin{aligned}&x_{L}=j \times 314 \times 0.0318 \Omega \\&x_{L}=j 9.9852 \Omega\end{aligned}$

The capacitive reactance is,

$\begin{aligned}x_{c} &=\frac{1}{j 2 \pi f c} \\x_{c} &=\frac{1}{j 2 \pi 314 \times 63.6 \times 10^{-6}} \\x_{c} &=-j 50.07 \Omega\end{aligned}$

i) The net impedance offered by the circuit is

$\begin{aligned}&z=R+j x_{L}-j x_{c} \\&z=10+j 9.9852-j 50.07 \\&z=10-j 40.08 \\&z=41.30 L-76^{\circ}\end{aligned}$

The expression of current is,

$\begin{aligned}i(t) &=\frac{e(t)}{2} \\i(t) &=\frac{100 \sin 314 t}{41.30 L-76^{\circ}} \\i(t) &=2.421 \sin \left(314 t+76^{\circ}\right) \mathrm{A}\end{aligned}$

ii) From the expression of the current the angle low voltage and current $p_{3} ,\\ \theta=76^{\circ} \\ \theta=76^{\circ}$ (capacitive)

iii) The power factor is given by the cosine of angle below the current of voltage

$P f=\cos \theta \\P f=\cos 76^{\circ} \\Y_{f}=0.242$

(iv) According to the question $V=200 \sin 314 t , R=10 \Omega$

comparing it with the original equation we get, $\mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$

(1) RMS voltage , $\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{m}}}{\sqrt{2}}=0.707 \times 200=141.4 \mathrm{~V}$

(2)RMS current , $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{R}}=\frac{141.4}{10}=14.14 \mathrm{~A}$

power dissipate,$\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \theta$

here$\cos \theta$ is power factor which is equal to 1 for pure resistive circuit since voltage and current are in same phase,

$\mathrm{p}=141.4 \times 14.14=1999.69 \simeq 2000 \mathrm{~W}$