Answer to Question #111038 in Electric Circuits for havefun7741

Question

[img]https://upload.cc/i1/2020/04/20/y3Yo4L.jpg[/img]


question is in photo . R=4

Answer 1

Given: $C=100\mu F; R=34\Omega; v(t)=220\cdot sin(314\cdot t)V$

Solution: (a) An AC voltage source in our problem has $w=314 rad/s$ and can be written as $v(t)=Im(V_a\cdot e^{iwt})$ where $V_a=220V$ (Note $e^{iwt}=cos(wt)+i\cdot sin(wt)$ ).

To find the current in a series AC circuit we determine the component of its impedance

(1) $Z=R+\frac{1}{iwC}=R-iX_c$ , where $X_c=\frac{1}{wC}=\frac{1}{100\cdot 10^{-6}F\cdot 314 rad/s}=31.8\Omega$

We can write the impedance in a form

(2) $Z=|Z|\cdot e^{i\phi}$ , where $|Z|=\sqrt{R^2+X_c^2}=\sqrt{34^2+31.8^2}\Omega=46.6\Omega$ and $\phi=arctan(\frac{-X_c}{R})=-arctan(\frac{31.8}{34})=-0.753 rad$ ($arctan$ also denote as $tan^{-1}$)

(2) According to the definition of impedance we have $Z=\frac{v}{i}$ thus

(3) $i=\frac{v}{Z}$ , where $v=V_a\cdot e^{iwt}$ and

(4) $i(t)=Im(i)=Im(\frac{V_a\cdot e^{iwt}}{|Z|e^{i\phi}})=\frac{V_a}{|Z|}\cdot Im(e^{iwt}\cdot e^{-i\phi})=\frac{V_a}{|Z|}sin(wt-\phi)$ and finally

(5) $i(t)=\frac{V_a}{|Z|}sin(wt-\phi)=\frac{220V}{46.6\Omega}sin(314t+0.753)=4.72\cdot sin(314t+0.753)A$

(b) Real power of our circuit is

$P=|I_{rms}|^2\cdot R=(\frac{4.72}{/\sqrt{2}})^2\cdot A^2\cdot 34\Omega=379W$

Reactive power is

$Q=|I_{rms}|^2\cdot X_c=(\frac{4.72}{/\sqrt{2}})^2\cdot A^2\cdot 31.8\Omega=354VAr$

apparent power is

$S=\sqrt{P^2+Q^2}=519 VA$

The power factor is $\frac{P}{S}=\frac{379}{519}=0.73$ or $73$%

(c) The average power consumed by the circuit is equals $P=378.7W$ as the reactive power isn't consumed.

Answers: (a) The current is $i(t)=4.72\cdot sin(314t+0.753)A$

(b) The power factor of the circuit is 73%

(c) The average power consumed by the circuit is 379 W