# Question with solving for complex power, S, for each component

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 :)

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 ${2\over \sqrt2}=\sqrt 2 V$.

• 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). Sep 2, 2014 at 0:04
• 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. Sep 2, 2014 at 5:46
• 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: $i(t)=r\cdot cos(2\pi f t+\phi)=1.97\cdot cos(10t+0.098)$. Sep 2, 2014 at 5:56

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 $S_{R_1}=\frac{19802}{10201}\, \mathrm{VA}$.

• 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? Sep 1, 2014 at 23:56
• @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. Sep 2, 2014 at 0:24
• @JTHiquet Oops, I flipped a sign, inductive reactive power is positive, capacitive reactive power is negative. Answer updated. Sep 2, 2014 at 9:03