# How to solve this task that relates to power in circuit?

Two impedance's are added together in parallel. $$\ Z_{1} = 2 - j5 (Ω) \$$ ,$$\ Z_{2} = 1 + j (Ω)\$$ Power on $$\ Z_1 = 20 W\$$. Determine the reactive power.

I tried this $$\P = U * I\$$ => $$\ P = \frac{U^2}{Z_1} \$$ and that would give me the absolute value of U = 10.38 V ,after that using voltage I would find the current that goes through both Impedance's. Following that logic I would then find the current of the source and multiply that with Impedance's ( $$\ 6.82\angle{-29.91} * 1.52\angle{29.93} = 61.33 - 35.25j \$$ ) and that would give me the wrong answer.

Is there any tip, I think my understanding of this whole concept of power in a circuit is a bit loose, so help about what I did wrong is greatly appreciated.

• Remember load is a pythagoras vector aka complex impedance and Power is the i real axis for R and j is the reactive axis for X. So use Trig. to find X – Tony Stewart Sunnyskyguy EE75 Jul 28 '19 at 15:21
• Maybe I was not clear sorry about that, In my language power means apparent power, real power means real power and reactive power well reactive power – Gustav Robert Kirchhoff Jul 28 '19 at 15:28
• Multiplying polar quantities means multiply the magnitudes and add the phase angles. 6.82 L-29.91 x 1.52 L29.93 = 10.37 L0.02 – Chu Jul 28 '19 at 17:29

Is there any tip

If you know the power on Z1 then you know that that is dissipated by the "2" part of 2 - j5 and this leads on to be able to state the current through that 2 ohms: -

$$20 = I^2 \cdot 2$$

Hence current is $$\\sqrt{10}\$$ = 3.162 amps.

From that you can calculate the line voltage using Z1's impedance. That impedance is $$\\sqrt{2^2+5^2}\$$ = 5.285 ohms. So you have the line voltage of 17.03 volts.

Can you take it from here?

In engineering, power is real (or active) power, apparent power is volts x amps and, reactive power is $$\\sqrt{(V\cdot I)^2 - (watts)^2}\$$: -