# How to find True power dissipated and apparent power?

I have to find the true power dissipated and apparent power for this RLC AC circuit. I've already found the impedance, the current from the supply and the current from each branch but I've forgotten how to find the true power dissipated and apparent power. Help would be appreciated. Sorry for the rough circuit.

• (1) What is the power dissipated in R? Edit your question and show your work. Commented Oct 31, 2020 at 9:29
• I haven't got a scanner to post my working. Is the power dissipated in R just Voltage x the current from the branch? Commented Oct 31, 2020 at 9:38
• You don't even need to work out the current. You can express power as a function of V and R only. Start with $P = VI$ and get rid of $I$ using Ohm's Law. Just type the answer in. That part is only two lines. Commented Oct 31, 2020 at 9:43
• so 240Vx3.259A (current in R branch) = 846.96W what do i do after this. Commented Oct 31, 2020 at 9:59
• Your answer is correct but I was prompting you to derive the formula $P = \frac {V^2} R$ which avoids you having to calculate the current. Have a read of allaboutcircuits.com/textbook/alternating-current/chpt-11/… and see if you can work out the others. Remember that the reactive powers will be opposite for the inductor and capacitor so they will tend to cancel out. (This is the principle of power-factor correction.) Commented Oct 31, 2020 at 10:12

I've forgotten how to find the true power dissipated and apparent power.

You've already calculated true power in comments but not so accurately. True power is (as transistor points out) $$\V^2/R\$$ = 847.06 watts and, the current in the resistor is 3.5294 amps.

For the apparent power I'd calculate individually the currents in the inductor and capacitor, these being: -

• Inductor current = 15.279 amps
• Capacitor current = 4.2977 amps

And, the trick here is to realize that that actual current that flows into the pair of them is the difference current. This is because L and C form a partial tuned circuit.

• Net LC current is 10.981 amps

Total current is resistive plus net reactive current added as phasors (or using pythagoras): -

$$\sqrt{3.5294^2+10.981^2}$$

And that equals 11.535 amps (this is the total RMS current from the 240 volt supply).

Hence apparent power is 240 x 11.535 = 2768.3 watts.

Well, the input impedance is given by:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{\frac{1}{\text{R}}+\frac{1}{\text{j}\omega\text{L}}+\text{j}\omega\text{C}}\tag1$$

The true power is given by:

$$\text{P}=\text{V}_{\text{in}}^*\cdot\text{I}_{\text{in}}^*\cdot\cos\left(\varphi\right)\tag2$$

Where $$\x^*\$$ is the RMS value.

In your case, $$\\text{V}_{\text{in}}^*=240\$$ is given. So, we get:

• $$\text{I}_{\text{in}}^*=\frac{1}{\sqrt{2}}\cdot\frac{240}{\left|\underline{\text{Z}}_{\space\text{in}}\right|}\tag3$$
• $$\varphi=\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)\tag4$$

It is not hard, to show that:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{68 \left(\left(\frac{1}{5 \pi }-\frac{57 \pi }{10000}\right)^2+\frac{1}{4624}\right)}+\frac{2890000 \pi \left(57 \pi ^2-2000\right)}{59642000 \pi ^2-938961 \pi ^4-1156000000}\cdot\text{j}\tag5$$

So:

• $$\text{I}_{\text{in}}^*=\frac{3 \sqrt{578000000-29821000 \pi ^2+\frac{938961 \pi ^4}{2}}}{2125 \pi }\tag6$$
• $$\varphi=-\arctan\left(\frac{68}{5\pi}-\frac{969\pi}{2500}\right)\tag7$$

$$\text{P}=\frac{7200 \sqrt{2}}{17}\approx598.961\space\text{W}\tag8$$