# RL Circuit Power Computation

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1. Mar 10, 2017

### peroAlex

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.

1. The problem statement, all variables and given/known data
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:
2. Relevant equations
I know that real power is computed as $$P = I^2 R$$ and reactive power as $$Q = I^2 X$$

3. 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/

2. Mar 10, 2017

### Staff: Mentor

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").