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In summary, complex numbers come into the game only because of mathematical convenience. All the quantities in AC circuit theory are, of course, real. The reason to use complex numbers is that it is easier to deal with exponential functions than trigonometric functions, since we consider an AC with a single frequency ##f=\omega/(2 \pi)##. E.g., a voltage is described as ##U(t)=U_0 \cos(\omega t)##, which is a real quantity. Now you can also describe this as ##U(t)=\text{Re}[U_0 \exp(\mathrm{i} \omega t)]##.To describe a circuit youf

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Complex numbers come into the game only because of mathematical convenience. All the quantities in AC circuit theory are, of course, real. The reason to use complex numbers is that it is easier to deal with exponential functions than trigonometric functions, since we consider an AC with a single frequency ##f=\omega/(2 \pi)##. E.g., a voltage is described as ##U(t)=U_0 \cos(\omega t)##, which is a real quantity. Now you can also describe this as ##U(t)=\text{Re}[U_0 \exp(\mathrm{i} \omega t)]##.

To describe a circuit you have to use differential equations. E.g., take a capacitor and a resistor in series with a voltage source. Then Faraday's Law, integrated along the circuit tells you that

$$Q/C+R \dot{Q}=U(t),$$

where ##Q## is the momentary charge on the capacitor. To solve this equation it's more simple to use the complex description with the exponential function for the voltage. Since the ##C## and ##R## are real you get then from any complex solution for ##Q(t)## the corresponding real solution by taking the real part.

To solve the equation note that you need the complete solution of the homogeneous equation (i.e., setting the right side to ##0##), which is

$$Q_{\text{hom}}(t)=A \exp[-t/(RC)],$$

where ##A## is an arbitray integration constant, and one special solution of the inhomogeneous equation. This we get with the ansatz

$$Q_{\text{inh}}(t)=B \exp(\mathrm{i} \omega t),$$

because the right-hand side is of this form, and taking derivatives always reproduces this exponential function. So plugging the ansatz into the equation, you get

$$\exp(-\mathrm{i} \omega t) B (1/C + \mathrm{i} \omega R)=U_0 \exp(-\mathrm{i} \omega t),$$

and this gives

$$B=\frac{U_0}{1/C+\mathrm{i} \omega R} = \frac{U_0 C}{1+\mathrm{i} \omega R C} = \frac{U_0 C (1-\mathrm{i} \omega RC)}{1+\omega^2 R^2C^2}.$$

So the complete solution is the sum of the homogeneous and the inhomogeneous solution

$$Q(t)=\frac{U_0 C (1-\mathrm{i} \omega R C)}{1+\omega^2 R^2 C^2} \exp(\mathrm{i} \omega t) + A \exp[-t/(R C)].$$

As you see, the homogeneous part goes exponentially to zero for ##t \rightarrow \infty##, i.e., after a sufficiently long time only the inhomogeneous solution is relevant.

The current for this stationary state is given by

$$I(t)=\dot{Q}_{\text{inh}}(t)=\frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \exp(\mathrm{i} \omega t).$$

To get the phase shift between current and voltage you have to write the prefactor in "polar form". The modulus is [EDIT: corrected in view of #8]

$$\left | \frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \right|=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}},$$

and the argument

$$\varphi=\arccos \left (\frac{\omega RC}{\sqrt{1+\omega^2 R^2 C^2}} \right).$$

So we can write

$$I(t)=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}} \exp[\mathrm{i}(\omega t+\varphi)],$$

i.e., ##\varphi## is the phase shift. The physical quantity is the real part of this expression, i.e., just the same but instead of the exp function the cos function,

$$I(t)=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}} \cos(\omega t+\varphi).$$

To describe a circuit you have to use differential equations. E.g., take a capacitor and a resistor in series with a voltage source. Then Faraday's Law, integrated along the circuit tells you that

$$Q/C+R \dot{Q}=U(t),$$

where ##Q## is the momentary charge on the capacitor. To solve this equation it's more simple to use the complex description with the exponential function for the voltage. Since the ##C## and ##R## are real you get then from any complex solution for ##Q(t)## the corresponding real solution by taking the real part.

To solve the equation note that you need the complete solution of the homogeneous equation (i.e., setting the right side to ##0##), which is

$$Q_{\text{hom}}(t)=A \exp[-t/(RC)],$$

where ##A## is an arbitray integration constant, and one special solution of the inhomogeneous equation. This we get with the ansatz

$$Q_{\text{inh}}(t)=B \exp(\mathrm{i} \omega t),$$

because the right-hand side is of this form, and taking derivatives always reproduces this exponential function. So plugging the ansatz into the equation, you get

$$\exp(-\mathrm{i} \omega t) B (1/C + \mathrm{i} \omega R)=U_0 \exp(-\mathrm{i} \omega t),$$

and this gives

$$B=\frac{U_0}{1/C+\mathrm{i} \omega R} = \frac{U_0 C}{1+\mathrm{i} \omega R C} = \frac{U_0 C (1-\mathrm{i} \omega RC)}{1+\omega^2 R^2C^2}.$$

So the complete solution is the sum of the homogeneous and the inhomogeneous solution

$$Q(t)=\frac{U_0 C (1-\mathrm{i} \omega R C)}{1+\omega^2 R^2 C^2} \exp(\mathrm{i} \omega t) + A \exp[-t/(R C)].$$

As you see, the homogeneous part goes exponentially to zero for ##t \rightarrow \infty##, i.e., after a sufficiently long time only the inhomogeneous solution is relevant.

The current for this stationary state is given by

$$I(t)=\dot{Q}_{\text{inh}}(t)=\frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \exp(\mathrm{i} \omega t).$$

To get the phase shift between current and voltage you have to write the prefactor in "polar form". The modulus is [EDIT: corrected in view of #8]

$$\left | \frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \right|=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}},$$

and the argument

$$\varphi=\arccos \left (\frac{\omega RC}{\sqrt{1+\omega^2 R^2 C^2}} \right).$$

So we can write

$$I(t)=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}} \exp[\mathrm{i}(\omega t+\varphi)],$$

i.e., ##\varphi## is the phase shift. The physical quantity is the real part of this expression, i.e., just the same but instead of the exp function the cos function,

$$I(t)=\frac{U_0 \omega C}{\sqrt{1+\omega^2 R^2 C^2}} \cos(\omega t+\varphi).$$

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It may be worth while pointing out that there is only a phase shift because of the finite resistances in a circuit. If you connect a capacitor across an AC voltage source (i.e. zero source resistance) the current flowing into the Capacitor will be totally in phase with the volts.how does capacitors and inductors cause phase difference between current and voltage?

When a series R is added, there is a finite time lag between the supply voltage and the instantaneous charge in the capacitor.

Edit Warning - this is rubbish! My head was not screwed on right when I made this hurried post. It's the Charge on the C that's in step with the Volts.

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No. The capacitor current is shifted wrt the voltage by 90If you connect a capacitor across an AC voltage source (i.e. zero source resistance) the current flowing into the Capacitor will be totally in phase with the volts.

The resistor does create a phase shift in the capacitor voltage wrt the source voltage. There is no phase shift without it because there is only one voltage; it is in phase with itself.

- #5

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Since this thread is at the Basic level, I'm not sure that you can follow it without knowledge of basic calculus. Have you studied calculus yet? What about complex numbers, like polar and rectangular forms and Euler's formula? There are descriptions of this without calculus, but that's not how I learned it, and I suspect they aren't really accurate. Let us know if you need an explanation at that level.

Khan Academy has some really good free tutorials about AC circuits. I would check them out.

Then, definitely not Basic level, after you have studied EE/physics for a while you will learn about Laplace transforms. This is the tool EEs working in industry actually use because it makes the problem much simpler to calculate. This relies on the complex representation of impedance, voltages, and currents. But it's really the same thing as above.

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Whoops - you're right. q = Cv so differentiate both sides. . . . .No. The capacitor current is shifted wrt the voltage by 90o because i=Cdvdt .

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If you set ##R=0## in my treatment of the RC-series circuit you get ##\varphi=\pi/2##, i.e., the current is by 90 degrees advanced compared to the voltage (in the stationary state of course).It may be worth while pointing out that there is only a phase shift because of the finite resistances in a circuit. If you connect a capacitor across an AC voltage source (i.e. zero source resistance) the current flowing into the Capacitor will be totally in phase with the volts.

When a series R is added, there is a finite time lag between the supply voltage and the instantaneous charge in the capacitor.

- #8

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I did notice a small typographical error here.

The denominator on the right side should be under a square root.To get the phase shift between current and voltage you have to write the prefactor in "polar form". The modulus is

$$\left | \frac{U_0 C \omega (\omega R C+\mathrm{i})}{1+\omega^2 R^2 C^2} \right|=\frac{U_0 \omega C}{1+\omega^2 R^2 C^2},$$

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Thanks for finding the typo. I've corrected it (also in the following equations).

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