# Calculating the reactive and active power

This is probably an easy one for you. In the following circuit simulate this circuit – Schematic created using CircuitLab

I want to calculate the active and reactive power in load R2-C3. I've calculated the voltage over these elements to be

$$u=\hat{U} \sin(\omega t + \phi) = 0.0465 \sin(1000 t + 3.53 \text{ rad}) V$$

Now I calculate the active power, P, to be

$$P = \frac{U_e^2}{R2}$$

and the reactive power, Q, to be

$$Q = \frac{U_e^2}{\Im (Z_{C3})}$$

Where $$Z_{C3}$$ is the impedance corresponding to C3 and $$U_e = \frac{\hat{U}}{\sqrt{2}}$$ is the effective value, or root mean square, of voltage u.

Am I doing this correctly?

• Are you assuming that the phase angle of $U_e$ is zero or does it have a reactive component? – Elliot Alderson Dec 12 '18 at 16:29
• The input power comes from I1 and not a voltage source so how would you calculate the voltage? Plus, there is no statement about the transformer's turns ratio or parasitic components. – Andy aka Dec 12 '18 at 16:33
• See my edits. I've already dealt with the transformer and know the voltage over R2-C3. So it was probably confusing, and unnecessary, to include the previous circuit in the question. – user1176517 Dec 12 '18 at 16:45
• Take the voltage as the zero phase angle reference, then calculate $U^2/R$ for the power and $U^2/X$ for the reactive VA, where $U$ is the RMS value. – Chu Dec 12 '18 at 23:17

$$P=\frac{U_e^2}{R_{tot}}=\frac{U_e^2}{(R_2\parallel R_3)}=\frac{U_e^2}{\frac{R_2R_3}{R_2+R_3}}$$
$$P=\frac{U_e^2}{Z_{tot}}=\frac{U_e^2}{(R_2\parallel Z_{C_3})}$$
So, it's now a little more complicated to get $$\\Re(P)\$$ and $$\\Im(P)\$$, but that's just math...