RL Circuit Power Computation: What Am I Missing in My Solution?

In summary, the student is having problems understanding why their computation deviates from solutions. They state that they are unable to uncover how and where the mistake occurred, but their results show that the currents should have equal real components according to solutions. Power calculations turn to the worse when reality is not what they should be. They ask for help, but note that the image associated with the problem is not accessible to the public.
  • #1
peroAlex
35
4
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.

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/
 
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  • #2
Please upload your image so that members don't have to go off-site to see it. Right now the image is not accessible to the public (site says "You need permission").
 

Related to RL Circuit Power Computation: What Am I Missing in My Solution?

1. What is an RL circuit?

An RL circuit is an electrical circuit that consists of a resistor (R) and an inductor (L) connected in series. It is commonly used in electronic devices to control the flow of current and voltage.

2. How do you calculate the power in an RL circuit?

The power in an RL circuit can be calculated using the formula P = I²R, where P is power in watts, I is current in amperes, and R is resistance in ohms. This formula is applicable when the circuit is in steady state.

3. What is the difference between instantaneous power and average power in an RL circuit?

Instantaneous power is the power at a specific moment in time, while average power is the average of the instantaneous power over a given time period. In an RL circuit, the instantaneous power can fluctuate due to the changing current and voltage, while the average power remains constant.

4. How does the power in an RL circuit change over time?

In an RL circuit, the power initially increases as the current builds up, then decreases as the inductor stores energy. Once the current reaches its maximum value, the power decreases as the inductor releases energy and the current decreases.

5. Can the power in an RL circuit be negative?

Yes, the power in an RL circuit can be negative. This occurs when the current and voltage are out of phase, meaning they do not peak at the same time. In this case, the power is considered negative because the inductor is absorbing energy from the circuit instead of supplying it.

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