# Power factor in AC circuits

1. A lagging power factor means that it has an inductive load, right?
2. The final circuit will still be lagging meaning that it will still be inductive?
3. How is the power factor getting used here? I can get net impedence in the final circit and get the power factor from that impedence triangle, but how am I to use the 0.5 power factor?

• In one of your previous questions you said this as a reply to a comment I made: Yes i know inductive components make it lag and capacitive components make pf leading <-- so, it seems you already know the answer to (i) and (ii). Is it just (iii) that needs an answer? Mar 11, 2023 at 16:11
• Yes precisely so. I can't attack this question like the other ones. I don't know where to start. Mar 11, 2023 at 16:13
• I think you should take note what I said on my previous answer Mar 11, 2023 at 16:23
• Power factor is used different ways by different people, so while you're in school, PF is what your coursework says it is. From a practical perspective though, Watts is the power you actually draw. VA is the "current proportional to voltage" ideal sinewave that the utility must deliver to power your load. Power factor is the ratio, watts/VA. Mar 11, 2023 at 23:50

Capacitor improves power factor from $$\pf_1\$$ = 0.5 lag to $$\pf_2\$$ = 0.8 lag.

1. $$\pf_1 = \frac {P} {S_1}\$$, determine apparent power, $$\S_1\$$.

2. $$\S_1 = \sqrt {P^2 + Q_L^2}\$$, determine inductive reactive power, $$\Q_L\$$.

3. Use $$\pf_2\$$ to find new apparent power, $$\S_2\$$.

4. Use phythagoreus with $$\S_2\$$ and $$\P\$$ to find net reactive power, $$\Q_{NET}\$$.

5. $$\pf_2\$$ is still lagging, so $$\Q_{NET} = Q_L - Q_C\$$. Find $$\Q_C\$$.

6. Capacitor is added in parallel with load, so voltage is constant, use $$\Q_C = \frac {V_C^2} {X_C}\$$ to determine capacitive reactance $$\X_C\$$.

7. Use $$\X_C = \frac {1} {2 \pi f C}\$$ to determine capacitance.

With power factor correction, you have R and L in series, with C in parallel. This results in a series/parallel RLC circuit, but this analysis takes advantage of the power triangle, which is the same for series or series/parallel.

Do this:

• calculate how much the load takes now apparent power
• calculate how much the load takes now inductive reactive power; let it be Q1
• calculate how much the load should take apparent power in case the pf was lifted to 0.8, let the active power stay as it is. Let the new reduced apparent power be S2
• calculate how much the load should take inductive reactive power when the apparent power is reduced to S2. Let the new reduced inductive reactive power be Q2
• calculate the difference Q1-Q2.
• calculate which capacitor takes amount Q1-Q2 of capacitive reactive power.

There's one thing you did not tell: Is your system an one phase or a three phase system? In a three phase system you need actually 3 capacitors, each of them handles one third of the capacitive reactive power and the capacitance depends on the connection principle - is it vye or delta? That's because it affects the voltage over a capacitor.

Well, let's first solve for the values of $$\\text{R}\$$ and $$\\text{L}\$$ (which I assume are connected in series):

\begin{cases} \begin{alignat*}{1} \text{P}&=\text{V}_\text{eff.}\cdot\text{I}_\text{eff.}\cdot\cos\left(\varphi\right)\\ \\ \text{I}_\text{eff.}&=\frac{\text{V}_\text{eff.}}{\left|\underline{\text{Z}}_{\space\text{i}}\right|}=\frac{\displaystyle\text{V}_\text{eff.}}{\displaystyle\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\\ \\ \cos\left(\varphi\right)&=\cos\left(\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\right)=\cos\left(\arctan\left(\frac{\omega\text{L}}{\text{R}}\right)\right)=\frac{\displaystyle1}{\displaystyle\sqrt{1+\left(\frac{\omega\text{L}}{\text{R}}\right)^2}} \end{alignat*} \end{cases}\tag1

Where $$\\text{V}_\text{eff.}\$$ is the effective voltage across the load, $$\\text{I}_\text{eff.}\$$ is the effective current trough the load and $$\\cos\left(\varphi\right)\$$ is the PF of the load.

Given your values, we can see that:

\begin{cases} \begin{alignat*}{1} \text{I}_\text{eff.}&=\frac{\displaystyle\text{P}}{\displaystyle\text{V}_\text{eff.}\cos\left(\varphi\right)}\\ \\ \text{R}&=\frac{\displaystyle\left(\text{V}_\text{eff.}\cos\left(\varphi\right)\right)^2}{\displaystyle\text{P}}\\ \\ \text{L}&=\frac{\displaystyle\text{V}_\text{eff.}^2\cos\left(\varphi\right)\sqrt{1-\cos^2\left(\varphi\right)}}{\displaystyle\text{P}\omega} \end{alignat*} \end{cases}\tag2

Now, we can write down the equation for the PF, $$\\Delta_\text{C}\$$, when a capacitor is connected in parallel:

\begin{alignat*}{1} \Delta_\text{C}&=\cos\left(\arg\left(\left(\text{R}+\text{j}\omega\text{L}\right)\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}\right)\right)\\ \\ &=\cos\left(\arg\left(\frac{\displaystyle\left(\text{R}+\text{j}\omega\text{L}\right)\cdot\frac{1}{\text{j}\omega\text{C}}}{\displaystyle\text{R}+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\right)\right)\\ \\ &=\cos\left(\arg\left(\frac{\displaystyle\text{R}+\text{j}\omega\text{L}}{\displaystyle1-\omega^2\text{CL}+\omega\text{CR}\text{j}}\right)\right)\\ \\ &=\cos\left(\arg\left(\text{R}+\text{j}\omega\text{L}\right)-\arg\left(1-\omega^2\text{CL}+\omega\text{CR}\text{j}\right)\right)\\ \\ &=\cos\left(\arctan\left(\frac{\displaystyle\omega\text{L}}{\displaystyle\text{R}}\right)-\arg\left(1-\omega^2\text{CL}+\omega\text{CR}\text{j}\right)\right)\\ \\ &=\begin{cases} \cos\left(\arctan\left(\frac{\displaystyle\omega\text{L}}{\displaystyle\text{R}}\right)-\frac{\displaystyle\pi}{\displaystyle2}\right)&\text{if}\space1-\omega^2\text{CL}=0\\ \\ \cos\left(\arctan\left(\frac{\displaystyle\omega\text{L}}{\displaystyle\text{R}}\right)-\arctan\left(\frac{\displaystyle\omega\text{CR}}{\displaystyle1-\omega^2\text{CL}}\right)\right)&\text{if}\space1-\omega^2\text{CL}>0\\ \\ \cos\left(\arctan\left(\frac{\displaystyle\omega\text{L}}{\displaystyle\text{R}}\right)-\left(\frac{\pi}{2}+\arctan\left(\frac{\displaystyle\omega^2\text{CL}-1}{\displaystyle\omega\text{CR}}\right)\right)\right)&\text{if}\space1-\omega^2\text{CL}<0 \end{cases} \end{alignat*} \tag3

Where $$\\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$$.

Edit, solving $$\(3)\$$ for $$\\text{C}\$$ gives:

$$\text{C}=\frac{\text{P}\left(\Delta_\text{C}\sqrt{1-\cos^2\left(\varphi\right)}\pm\cos\left(\varphi\right)\sqrt{1-\Delta_\text{C}^2}\right)}{\Delta_\text{C}\cos\left(\varphi\right)\omega\text{V}_\text{eff.}^2}\tag4$$

So, for your case we get:

$$\text{C}=\frac{\displaystyle5000\left(\frac{4}{5}\cdot\sqrt{1-\left(\frac{1}{2}\right)^2}\pm\frac{1}{2}\cdot\sqrt{1-\left(\frac{4}{5}\right)^2}\right)}{\displaystyle\frac{4}{5}\cdot\frac{1}{2}\cdot2\pi\cdot50\cdot400^2}=\frac{4\sqrt{3}\pm3}{12800\pi}\tag5$$