- #1
peroAlex
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I'm having some problems understanding why my computation deviates from solutions. I thank in advance to anyone who can give me some tips where I've gone wrong.
Compute real power on resistors ## R_1 ## and ## R_2 ## and reactive power on inductor ## L ##. Circuit is powered by alternating current ## i(t) = 20 sin(\omega t) ## where ## \omega = 400 s^{-1} ##. Here's a link to picture associated with the problem:
https://drive.google.com/file/d/0ByeYlJxPvdrUUUx2RTVFbkF2SFk/view
I know that real power is computed as $$ P = I^2 R $$ and reactive power as $$ Q = I^2 X $$
Part where me and solutions are in phase (pun intended).
Since current distributes, we can write ## \dot{I_T} = \dot{I_1} + \dot{I_2} ##. Those are currents in phasor form. Total current ## \dot{I_T} ## is computed as shown: $$ i(t) = 20 sin(\omega t) = 20 cos(\omega t - \frac{\pi}{2}) \\ \text{phasor becomes } \dot{I_T} = -20j$$ Since voltage must be equal on both branches, we can write ## R_1 \dot{I_1} = (R_2 + j \omega L) \dot{I_2} ## and so solving this simple system of two equations yields $$ \dot{I_1} = \frac{16 \Omega + j 12\Omega}{36 \Omega + j 12 \Omega} \ \ \ \ \ \ \ \ \ \dot{I_2} = \frac{20 \Omega}{36 \Omega + j 12 \Omega}$$
From here on, my path separates from solutions.
It might be a mathematical mistake for which I'm unable to uncover how and where it occurred, but ## \frac{16 \Omega + j 12\Omega}{36 \Omega + j 12 \Omega} \frac{\text{/} \cdot 36 \Omega - j 12 \Omega}{\text{/} \cdot 36 \Omega - j 12 \Omega} ## results in ## \Re(I_1) = 0.3 ## and ## \Im(I_2) = 0.1 ##. Vice versa, ## \frac{20 \Omega}{36 \Omega + j 12 \Omega} \frac{\text{/} \cdot 36 \Omega - j 12 \Omega}{\text{/} \cdot 36 \Omega - j 12 \Omega} ## should mean that ## \Re(I_2) = 0.5 ## and ## \Im(I_2) = 0.166 ##.
However, according to solutions, both currents have equal real component ## I_1 = I_2 \approx 10.54 A ##.
From here on, power computation takes turn to the worse, since my results yield ## P_1 = \frac{1}{2} R_1 \Re(I_1)^2 = 0.9 W ##, ## P_2 = 2 W ## and ## Q = \frac{1}{2} \omega L \Re(I_2)^2 = 1.5 W ## when reality it should be $$ P_1 \approx 1111.11 W \\ P_2 \approx 888.88 W \\ Q \approx 666.66 V A $$
As said above, I would like to ask for some help on that one. Is it mathematical fault or am I missing an additional step? I am grateful for any help/
Homework Statement
Compute real power on resistors ## R_1 ## and ## R_2 ## and reactive power on inductor ## L ##. Circuit is powered by alternating current ## i(t) = 20 sin(\omega t) ## where ## \omega = 400 s^{-1} ##. Here's a link to picture associated with the problem:
https://drive.google.com/file/d/0ByeYlJxPvdrUUUx2RTVFbkF2SFk/view
Homework Equations
I know that real power is computed as $$ P = I^2 R $$ and reactive power as $$ Q = I^2 X $$
The Attempt at a Solution
Part where me and solutions are in phase (pun intended).
Since current distributes, we can write ## \dot{I_T} = \dot{I_1} + \dot{I_2} ##. Those are currents in phasor form. Total current ## \dot{I_T} ## is computed as shown: $$ i(t) = 20 sin(\omega t) = 20 cos(\omega t - \frac{\pi}{2}) \\ \text{phasor becomes } \dot{I_T} = -20j$$ Since voltage must be equal on both branches, we can write ## R_1 \dot{I_1} = (R_2 + j \omega L) \dot{I_2} ## and so solving this simple system of two equations yields $$ \dot{I_1} = \frac{16 \Omega + j 12\Omega}{36 \Omega + j 12 \Omega} \ \ \ \ \ \ \ \ \ \dot{I_2} = \frac{20 \Omega}{36 \Omega + j 12 \Omega}$$
From here on, my path separates from solutions.
It might be a mathematical mistake for which I'm unable to uncover how and where it occurred, but ## \frac{16 \Omega + j 12\Omega}{36 \Omega + j 12 \Omega} \frac{\text{/} \cdot 36 \Omega - j 12 \Omega}{\text{/} \cdot 36 \Omega - j 12 \Omega} ## results in ## \Re(I_1) = 0.3 ## and ## \Im(I_2) = 0.1 ##. Vice versa, ## \frac{20 \Omega}{36 \Omega + j 12 \Omega} \frac{\text{/} \cdot 36 \Omega - j 12 \Omega}{\text{/} \cdot 36 \Omega - j 12 \Omega} ## should mean that ## \Re(I_2) = 0.5 ## and ## \Im(I_2) = 0.166 ##.
However, according to solutions, both currents have equal real component ## I_1 = I_2 \approx 10.54 A ##.
From here on, power computation takes turn to the worse, since my results yield ## P_1 = \frac{1}{2} R_1 \Re(I_1)^2 = 0.9 W ##, ## P_2 = 2 W ## and ## Q = \frac{1}{2} \omega L \Re(I_2)^2 = 1.5 W ## when reality it should be $$ P_1 \approx 1111.11 W \\ P_2 \approx 888.88 W \\ Q \approx 666.66 V A $$
As said above, I would like to ask for some help on that one. Is it mathematical fault or am I missing an additional step? I am grateful for any help/
