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schematic

simulate this circuit – Schematic created using CircuitLab

Hi!

So I have a couple questions that I have already tried to solve that I think I am going in a little bit of the wrong direction with.

So at first I am suppose to find the source current \$i\$:

What I have tried to do seemed too simple. I made it so \$t=0\$. And so the voltage source was \$2\$ V. Then I did a KVL and got \$i = 1\$A. But I feel like I should've used impedances.

I know the impedances: Capacitor = \$-10j\$ Inductor = \$10j\$ Resistors = \$1\$

How do I find the complex power of each component with the impedences?

Also side noting:

How do I change \$2\cos(10t)\$ to Vrms?

Thank you :)

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2 Answers 2

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1) The total impedance of your circuit is

$$Z_{total}=R_1+(-j{{1}\over{10 C_1}}||j\cdot 10L_1||R_2)=R_1+{1 \over j\cdot (10C_1-{1 \over 10L_1})+{1 \over R_2} }=1+{1 \over 1+9.9j} \approx 1.01+j\cdot 0.0999 \Omega \approx 1.01504 \cdot e^{-j\cdot 0.098}\Omega$$

From that, you can easily calculate the complex amplitude of the current:

$$I={V_1 \over Z_{total}}={2\over1.01504 \cdot e^{-j\cdot 0.098}}\approx 1.97\cdot e^{j\cdot 0.098} A$$

2) The complex power through a resistor, capacitor or inductor can be expressed as \${V\cdot I^*} \over 2\$, where \$V\$ is the complex amplitude of the voltage across the element and \$I^*\$ is the conjugate of the complex amplitude of the current flowing through it.

  • The current of \$R_1\$ is the current calculated in 1). The voltage across it can be expressed as \$V_{R1}=I\cdot R_1=1.97\cdot e^{j\cdot 0.098}V\$. Therefore, \$S_{R1}={1.97\cdot e^{j\cdot 0.098}\cdot1.97\cdot e^{-j\cdot 0.098}\over2}={1.97^2\over 2}=1.94045VA\$. It is a resistor, so its power is purely real. (By the way, the ideal reactive elements – capacitors, inductors – have purely reactive power.)
  • Let the impedance of the 3 parallel elements be \$Z_2\$. The voltage across them is common, let's call it \$V_2=V_1\cdot {Z_2\over R_1+Z_2}={Z_2 \over Z_{total}}\$ (voltage division between \$R_1\$ and the parallel elements). I'm not going through the calculations, but every power can be now expressed as \${V_2\cdot I_{element}^* \over 2}={V_2^2 \over Z_{element}}\$.

3) Side note: The RMS value of a sine/cosine function is calculated as \$amplitude\over \sqrt 2\$, in our case it is \${2\over \sqrt2}=\sqrt 2 V\$.

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  • \$\begingroup\$ I dont understand how you jumped to a conclusion in your impedance total. Where did the e^-0.098j come from? How did you transition the impedance to a numeric result like that? I got mine to equal the same as the answer below this.. Which is 10j/(10j-99). \$\endgroup\$
    – JT Hiquet
    Sep 2, 2014 at 0:04
  • \$\begingroup\$ That is not the total impedance, only the impedance for the parallel RLC part. You still need to add the value of \$R_1\$, which is in series with them. \$\endgroup\$
    – hryghr
    Sep 2, 2014 at 5:46
  • \$\begingroup\$ Regarding your question: complex numbers have several different interpretations, one being \$a+bj\$, where \$a\$ is the real and \$b\$ is the imaginary part. Another one is \$r\cdot e^{j\phi}\$, where \$r\$ is the amplitude and \$\phi\$ is the phase. (You can get the second parameters from the first ones like this: \$r=\sqrt{a^2+b^2}\$ and \$\phi=arctg({b\over a})\$.) The amplitude-phase representation helps calculate time-domain functions: \$i(t)=r\cdot cos(2\pi f t+\phi)=1.97\cdot cos(10t+0.098)\$. \$\endgroup\$
    – hryghr
    Sep 2, 2014 at 5:56
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To get complex power, you first need to get voltage on the three components after R1. You can calculate it like a voltage divider, when you first combine impedances of R2,C1,L1. \$Z_{L_1}=j\omega L_1=10j\Omega\$, \$Z_{C_1}=\frac{1}{j\omega C_1}=-0.1j\Omega\$, \$Z_{R_2}=1\Omega\$. Make inverse terms to get admittance, add them and invert again. \$Z_{R_2,C_1,L_1}=\frac{10j}{10j-99}\Omega\$. RMS voltage of source is \$\frac{2}{\sqrt{2}}\mathrm{V}\$, scaled by \$\dfrac{\frac{10j}{10j-99}}{\frac{10j}{10j-99}+1}\$ it is \$\frac{200\sqrt{2}}{10201}-\frac{990\sqrt{2}}{10201}i\, \mathrm{V}\$.

Now you can start with power calculation. \$S=V\cdot I^*=\frac{V\cdot V^*}{Z^*}\$, \$S_{L_1}=\frac{20}{10201}i\, \mathrm{VA}\$, \$S_{C_1}=-\frac{2000}{10201}i\, \mathrm{VA}\$, \$S_{R2}=\frac{200}{10201}\, \mathrm{VA}\$

Last thing is the power at R1. Voltage at it is \$\frac{2}{\sqrt{2}} \cdot \dfrac{1}{\frac{10j}{10j-99}+1}=\frac{10001\sqrt{2}}{10201}+\frac{990\sqrt{2}}{10201}i\, \mathrm{V}\$, and \$S_{R_1}=\frac{19802}{10201}\, \mathrm{VA}\$.

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  • \$\begingroup\$ So i follow what you did until you did the power calculation. On SL1, how did you get -20/10201 i? and why is there an i? \$\endgroup\$
    – JT Hiquet
    Sep 1, 2014 at 23:56
  • \$\begingroup\$ @JTHiquet Voltage is \$\frac{200\sqrt{2}}{10201}-\frac{990\sqrt{2}}{10201}i\$, its complex conjugate is \$\frac{200\sqrt{2}}{10201}+\frac{990\sqrt{2}}{10201}i\$. Their product is \$\frac{200}{10201}\$. And divided by \$10i\$ it is \$-\frac{20}{10201}\$. The i and negative sign denote it is inductive reactive power. To be exact, it probably should be marked \$\widehat{S_{L_1}}\$, like a phasor. \$\endgroup\$
    – venny
    Sep 2, 2014 at 0:24
  • \$\begingroup\$ @JTHiquet Oops, I flipped a sign, inductive reactive power is positive, capacitive reactive power is negative. Answer updated. \$\endgroup\$
    – venny
    Sep 2, 2014 at 9:03

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