# How to use phasor algebra to solve for capacitive reactance?

I have an RLC series circuit.

Voltage supplied by volatage source as a function of time: $V(t) = 230 \sin(\omega t+\pi/4)A$.

Current in circuit as a function of time: $I(t)=10\sin(\omega t - \pi/6)V$

The value of the resistance is $5\Omega$ and value of inductive reactance is $8j\Omega$. I need to find the value of capacitive reactance $X_{C}$.

I tried to solve like this (using phasors i.e. polar representation):

$$230/\sqrt{2}e^{j\pi/4} = (5 + 8j + X_{C})(10/\sqrt{2}e^{-j\pi/6})$$ $$\implies 23e^{j(\pi/4+\pi/6)}-5-8j =X_{C}\implies X_{C}=(0.95+14.21j)$$

[Ohm's Law]

However, while writing this, I realized that the left hand side's real part is not equal to the right hand side's real part. When I solve for $X_{C}$ I get $(0.95+14.21j)\Omega$ which is impossible since $X_{c}$ (capacitive) must be imaginary with a phase factor of $-j$.

I'm confused about how to use phasor algebra to solve this problem. Any help will be appreciated.

• You should make your series impedance: $Z=\sqrt{R^2+(X_L-X_C)^2}$ ... Now solve for $X_C$.
– user103380
Commented Mar 8, 2018 at 15:48
• ALSO: I must ask, when you put V(t) and I(t), did you consider that it's an RMS value? Don't forget to divide by $\sqrt{2}$ so it should be $230e^{j\pi/4}/\sqrt{2}$.
– user103380
Commented Mar 8, 2018 at 15:55
• @KingDuken Can you please explain why my method of using Ohm's law is wrong? Why isn't the real part same on both left and right?
– user133614
Commented Mar 8, 2018 at 15:55
• @KingDuken I think that the $\sqrt{2}$ factor gets cancelled anyway in the Ohm's law expression
– user133614
Commented Mar 8, 2018 at 15:56
• When you do Ohm's Law, you have to consider your R,L, and C values as a single impedance. Also, I believe you have impedance confused with reactance as when you calculate reactance, you already consider the imaginary number "j". Therefore, when you solve for reactance, there is no "j" in the formula.
– user103380
Commented Mar 8, 2018 at 16:00

This is how I solved, taking cos as reference to phasor equations: $$V(t) = 230 \sin(\omega t+\pi/4)$$ $$I(t)=10\sin(\omega t - \pi/6)V$$ let $(\omega t+\pi/4) = \phi$

we know: $$sin\phi = -cos(\pi/2+\phi)$$ $$\implies V(t) = -230cos(\omega t+3\pi/4)$$ $$I(t) = -10cos(\omega t+\pi/3)$$ Therefore in phasor form, it V and I can be represented as: $$V = -230 e^{j3\pi/4}$$ $$I = -10e^{j\pi/3}$$ As everything is in phasor form, now we can apply ohm's law directly:

$$230e^{j3\pi/4} = (5 + 8j + X_{C})(10e^{j\pi/3})$$ $$\implies 23 e^{j5\pi/12} = (5 + 8j + X_{C})$$ $$\implies X_c = 23 cos(5\pi/12) + j 23 sin(5\pi/12) -5 - 8j$$ $$= 5.95 + 22.21 j - 5 -8j$$ $$= 0.95 + 14.21j$$ This will satisfy the given conditions in the question. As we can see the obtained expression for $X_c$ has a real part or resistive part and a positive imaginary part or inductive reactance part. Pure capacitance will only have negative imaginary part, i.e, with phasor -90 degrees.

• $23\cos(5\pi/12)$ is around $5.95$. Please check
– user133614
Commented Mar 8, 2018 at 18:11
• No he didnt take into account the phase difference between V and I given in the question. If we imagine a pure capacitive reactance of 30.45 ohms, it will satisfy the currents value 10 A. But it will violate the phase given in the question. if you calculate the phase with 30.45, you will obtain it as $13.69\pi/100$. But in the question, it is already given as $\pi/4 + \pi/6 = 5\pi/12$ Commented Mar 8, 2018 at 18:22
• Anyway, in your last step there will be a plus sign. $22.21-8=+14.21$
– user133614
Commented Mar 8, 2018 at 18:33
• actually, $sin(\theta)=cos(\theta -\frac{\pi}{2})$. Still works in this case because you shift the voltage and current the same amount, the negative signs cancel.
– Big6
Commented Mar 8, 2018 at 19:25
• Also strictly speaking, you found an impedance, since you have three components in series, you add up the 'impedances', (you didn't solve for an actual reactance). Now, the impedance you found does not represent a capacitor, as you proved, it could be modeled as a resistor plus some inductance. Not taking anything away from your answer: The idea you proved still holds!
– Big6
Commented Mar 8, 2018 at 19:31

Any help will be appreciated.

You are supplying $\frac{230}{\sqrt2}$ volts and the load is taking $\frac{10}{\sqrt2}$ amps.

That's an overall impedance of 23 ohms.

You also know that 5 ohms is purely resistive hence: -

The reactive impedance is $\sqrt{23^2-5^2}$ = 22.45 ohms

Given that 8 ohms of the reactance is inductive and that capacitve and inductances are opposing, to get 22.45 ohms net, the capacitve reactance must be 30.45 ohms.

I'm confused about how to use phasor algebra to solve this problem.

Just in case you think I didn't use phasor algebra, think about the $\sqrt{23^2-5^2}$ bit and how this relates to it.

• I think your method is perfect. But could you please explain why the real parts of the equation I wrote aren't same? Why is my method wrong, of solving for $X_{C}$?
– user133614
Commented Mar 8, 2018 at 16:02
• So, maybe I should think that any help isn't appreciated. I showed you a method that is fundamentally based on phasor algebra as per your question and I really can't see why if you knew this method why you are not using it. Commented Mar 8, 2018 at 16:45
• Either there's a typo or the problem should be binned. Nominating components as R, L, and C is unambiguous and does not allow other arbitrary components to lurk in the shadows.
– Chu
Commented Mar 8, 2018 at 21:32
• I saw your comment in the accepted answer. There's actually a discrepancy between what the circuit shows and what's really in the place of the capacitor. Your approach is spot on but it hinges on the premise that there is a capacitor there. If you solve for the unknown impedance (what one thinks is Zc), it has the a + j*b form, so it looks more like a resistor in series with an inductor. There seems to be a mismatch between the current/voltage, and phase given in the statement, it doesn't hold for a capacitor. Your approach is fine and would have worked...just a wrong premise...
– Big6
Commented Mar 9, 2018 at 1:03
• Yea. Also the 77.4 degree phase is negative. while in question its +75 degrees. Commented Mar 9, 2018 at 12:21