# How is the reactance of a capacitor formulated?

Question: Understanding the Reactance of a Capacitor

Hi everyone,

I'm having trouble formulating the reactance in a capacitor and would appreciate some help. Here's what I understand so far:

1. We have an alternating current as the voltage source, and a capacitor in the circuit.
2. The voltage input is given by $$\ V_{in} = V_0 \sin(\omega t) \$$.
3. The capacitance is defined as $$\ C = \frac{Q}{V} \$$.

Given this, the potential charge stored in the capacitor's electric field will oppose the change in voltage, which is the reactance.

Starting from the charge equation: $$Q(t) = C \cdot \frac{d(V_0 \sin(\omega t))}{dt} = C \cdot V_0 \cdot \omega \cdot \cos(\omega t)$$

The reactance $$\ X_c \$$ is defined as: $$X_c = \frac{V}{I} = \frac{\sin(\omega t)}{C \omega \cos(\omega t)}$$

I'm aware of Euler's formula: $$\ e^{j\theta} = \cos(\theta) + j\sin(\theta) \$$.

When I see the substitutions being made, people often only consider a part of Euler's formula. For example, in the above equation, $$\ \cos(\omega t) \$$ is replaced by something like $$\ \text{Re}(e^{j\omega t}) \$$. However, $$\ e^{j\theta} \$$ has both real and imaginary components. If we replace just a portion of it in our transformation, it seems like an imbalanced equation because we cannot get back to the original form.

Can someone shed some light on this? How exactly does the reactance formula $$\ X_c = \frac{1}{\omega C} \$$ come about?

Thank you!

• Starting from the charge equation: the equation you have written is for the current, $i_C(t)$, not for the charge. So it must have been $I(t) = ...$. Commented Jul 15 at 15:09

## The math

Consider this linear differential equation with real coefficients

$$\frac{\text{d}\ y(t)}{\text{d}t} + a_0y(t) = b_0x(t) \tag1$$

Assume now that the input and output are complex valued functions, with a real part and an imaginary part: $$\x(t) = x_R(t) + jx_I(t)\$$ and $$\y(t) = y_R(t) + jy_I(t) \$$. Inserting this in $$\(1)\$$ gives

$$\frac{\text{d}\ \Big(y_R(t) + jy_I(t) \Big)}{\text{d}t} + a_0\Big(y_R(t) + jy_I(t) \Big) = b_0 \Big(x_R(t)+jx_I(t) \Big) \tag2$$

$$\color{blue}{\frac{\text{d}y_R(t)}{\text{d}t}} + j\color{green}{\frac{\text{d}y_I(t)}{\text{d}t}}+\color{blue}{a_0y_R(t)}+j\color{green}{a_0y_I(t)} = \color{blue}{b_0x_R(t)} + j\color{green}{b_0x_I(t)} \tag3$$

Grouping the real and imaginary terms gives

$$\color{blue}{\bigg(\frac{\text{d}y_R(t)}{\text{d}t} + a_0y_R(t) \bigg)} + j \color{green}{\bigg(\frac{\text{d}y_I(t)}{\text{d}t} + a_0y_I(t) \bigg)} = \color{blue}{b_0x_R(t)} + j\color{green}{b_0x_I(t)} \tag4$$

Because the coefficients are real it is implied that

$$\color{blue}{\frac{\text{d}y_R(t)}{\text{d}t} + a_0y_R(t)} = \color{blue}{b_0x_R(t)} \\\\\\ j\color{green}{\bigg(\frac{\text{d}y_I(t)}{\text{d}t} + a_0y_I(t) \bigg)} = j\color{green}{b_0x_I(t)} \tag5$$

Conclusion:

• The real part of the system's output is only influenced by the real part of the input signal.
• The imaginary part of the system's output is only influenced by the imaginary part of the input signal

## Consequences

Even if we just want to find the system's response to a cosine function, we can define the input signal as being a complex exponential. The output will be a scaled and rotated complex exponential, and it's real part will be the response to the cosine function. For a capacitor we have

$$I_C(t) = C\frac{\text{d} \ V_C(t)}{\text{d}t} \tag6$$

For a cosine function as input, we have just established that defining $$\V_C(t)\$$ as a complex exponential and taking the real part would give the same result as if we only considered $$\V_C(t) = \cos(\omega t)\$$. Let's define $$\V_C(t) = e^{j\omega t}\$$ which yields when inserting in $$\(6)\$$

$$I_C(t) = j\omega C e^{j\omega t} \tag7$$

Taking the real part on the right-hand side would give us $$\I_C(t) = -\omega C \sin(\omega t) \$$, but let's instead carry on. Remember that $$\V_C(t) = e^{j\omega t}\$$ which causes $$\(7)\$$ to become $$I_C(t) = j\omega CV_C(t) \tag8$$ $$Z_C = \frac{V_C(t)}{I_C(t)} = \frac{1}{j\omega C} \tag9$$

$$\Z_C\$$ is the impedance of a capacitor and allows us to calculate the resulting current for any sinusoidal input voltage, by dividing the voltage with the impedance and taking the real part.

In general, impedance has a real part called resistance and an imaginary part called reactance, defined as

$$Z = R + jX \tag{10}$$

In the case for an ideal capacitor, the resistance is zero, and the reactance is

$$X_C = \text{Im} \Big[Z_C \Big] = \text{Im} \Big[ \frac{1}{j\omega C} \Big] = \text{Im} \Big[\frac{-j}{\omega C} \Big] = -\frac{1}{\omega C} \tag{11}$$

• Thank you, @Carl! This is good and what i was looking for!. It is fascinating that if a function has real and imaginary parts, plugging it into a linear differential equation (LDE) shows a relationship where the real part affects the original output, but the imaginary part has no effect. You did show me something, and I am happy to mark it as answered. Out of curiosity, does this only work for differential equations? You do not need to explain but just point me to a resource. I will learn from it. Btw Im lost with "we can define the input signal as being a complex exponential" went right over
– DPV
Commented Jul 15 at 16:32
• @DPV As far as I know this is only valid for linear differential equation with real coefficients. I could be wrong. Unfortunately, I can't point you to a resource - I don't have any. This is stuff I learned from my signals and linear systems in continuous time class years ago. The book I used in that course was "Signal processing and linear systems" by Lathi, which is the best textbook I've used for any subject, but it doesn't go into the depth of responses to real/imaginary components as I do here. A complex exponential is simply $e^{j\omega t}$ - a convenient function for solving ODEs
– Carl
Commented Jul 15 at 17:06
• Thank you, I appreciate it! @Carl
– DPV
Commented Jul 15 at 17:27

How exactly does the reactance formula Xc=1ωC come about?

$$\\mathbb{V}=V_0\exp\left(j\,\omega t\right)\$$. (Note: The vector length is unchanging. It's tip forms a helix traveling in time. A voltage plane slices the helix to create the real part. So $$\V_t= V_0\sin\left(j\,\omega t\right)\$$ doesn't function for these purposes.)

Then $$\\mathbb{I}=C\frac{\text{d}}{\text{d}t} \mathbb{V}=C\,j\omega\,V_0\exp\left(j\,\omega t\right)\$$.

To start:

\require{cancel}\begin{align*}{\mathbb{Z}_\text{C}}=\frac{\mathbb{V}}{\mathbb{I}}&=\frac{\cancel{V_0}\exp\left(j\,\omega t\right)}{C\:j\,\omega\,\cancel{V_0}\exp\left(j\,\omega t\right)}&&=\frac{\cancel{V_0}\cancel{\exp\left(j\,\omega t\right)}}{C\:j\,\omega\,\cancel{V_0}\cancel{\exp\left(j\,\omega t\right)}}\\\\&=\frac{\exp\left(j\,\omega t\right)}{\omega\,C\,j\exp\left(j\,\omega t\right)}&&=\frac1{j\,\omega\,C}\\\\&=\frac1{\omega\,C}\cdot\frac{\exp\left(j\,\omega t\right)}{j\exp\left(j\,\omega t\right)}&&=\frac1{\omega\,C}\cdot\frac{1}{j}\\\\&={\text{X}_\text{C}}\cdot\frac{\exp\left(j\,\omega t\right)}{j\exp\left(j\,\omega t\right)}&&={\text{X}_\text{C}}\cdot\frac1{j}\\\\&={\text{X}_\text{C}}\cdot\exp\left(j\left[\omega t+\phi\right]\right)&&\therefore \frac1{j}=\exp\left(j\left[\omega t+\phi\right]\right)=-j\end{align*}

The only solution to $$\\frac1{j}=\exp\left(j\left[\omega t+\phi\right]\right)\$$ is where $$\\omega t+\phi=\frac32\pi\$$ (or any integer number of $$\2\pi\$$'s away from there.) As the $$\j\,\omega\$$ part is removed for phasors, the answer is $$\\phi=-90^\circ\$$.

The exact same process applies to inductance, giving $$\{\mathbb{Z}_\text{L}}=j\,\omega\,L=j\,{\text{X}_\text{L}}\$$. The solution for the inductor then gives $$\\phi=+90^\circ\$$.

• Thank you for the explanation. My concern is regarding how we represent waveforms using phasors. Phasor representations of sine and cosine differ by an angle, so they do not cancel out as you described. Please enlighten me, if you want me to refer to some concepts to understand this, please point me there I will try to learn.
– DPV
Commented Jul 15 at 3:53
• @DPV Phasors are a stripped down simplification. I talk a little about it here. But the essence is that phasors remove the time-dependent part, leaving only the phasor (phase angle) part. Euler's is simply divided up into two parts -- the time-independent and time-dependent parts -- and then, since it is assumed that $\omega$ is a constant that can be brought back in whenever it's needed, adding back time, then the simplification just tosses out the time-dependent part. It can always be pasted back on whenever needed. Commented Jul 15 at 4:03
• @DPV Plotting resistance on the x-axis and then using the y-axis to plot reactance, you find that inductance has a positive j while capacitance (because the j is in the denominator) has a -j after multiplying by $\frac{j}{j}$. So the X is just the imaginary magnitude of the Z for reactance. The resulting vector has an angle, which is given the name phasor. Just spend some time with the math on paper. It will dawn on you. Commented Jul 15 at 4:07
• @DPV Also look here, I suppose, where I discuss a reason for using conjugates when talking about current. But most of all because I show how a phasor occurs in the math. I thought you were asking about how that particular reactance fraction was developed. Instead, it seems you want to know exactly what a phasor is. I misunderstood you, I guess. That link will tell you. Commented Jul 15 at 4:13
• Thank you @periblepsis. I appreciate your providing resources. I will go through your comments, but to clarify, Yes, I am aware that we use Euler's representation to strip out time dependency and view it as the maximum amplitude, which is our radius, and the radians represent the phase. Using this information, we can get the signal at radians, which will be synonymous with the time. Now, issue is: how is ( \tan(\theta / \omega_c) ) becoming ( 1 / \omega_c )? Since (\sin / \cos) is (\tan). I know that (\sin(\pi/2 - \phi) = \cos(\phi)). There is a 90° phase shift.
– DPV
Commented Jul 15 at 6:52

In an ideal capacitor, there are no parasitic resistances so the real component to the impedance is just zero and the phase angle is -90 degrees or -j. It can also be written as Z = 0 - jωC.

In real capacitors of course there are parasitic inductances and resistances but those can also be modeled to whatever degree of accuracy you require.