Mathematics Involving in LCR Circuit

I will first review the mathematical topic Complex Number, and then I will use the concept of the complex number to solve the differential equation involving the RLC circuit.

Review of Complex Number

Generally, we write complex number as \Z =x+iy. We, electrical engineers, had already used the symbol I for current, so we use j for representing the imaginary term.

If you have a real axis and an imaginary axis, then the complex number is a point in the complex plane. The length of the line joining the origin and that point is \mid Z\mid and is called the magnitude of that complex number. The angle \phi made by that line with the real axis is called the phase of that complex number. We can write complex number as Z=\mid Z \mid e^{j \phi} .

Multiplication of Complex Number

Suppose you have two complex numbers, Z_{1}=\mid Z_{1} \mid e^{j \phi_{1}} and Z_{2}=\mid Z_{1} \mid e^{j \phi_{2}} . For multiplying two complex numbers, we multiply the magnitude and add the phase, so Z_{1}.Z_{2}=\mid Z_{1}\mid.\mid Z_{2} \mid e^{j(\phi_{1} + \phi_{2})} . The important thing is that if you multiply one complex number by another complex number, you scale the complex number and you rotate it. Draw it. It will make sense.

Complex Conjugate and Real Part of Complex number

Every complex number Z=x+jy is associated with another complex number Z^{*}=x-jy. Z^{*} is called complex conjugate of Z . If you add Z and Z ^{*} and do little algebra, you get x=\frac{Z+Z^{*}}{2} , which is the real part of the complex number Z .

LCR Circuit

Consider an LCR circuit, excited by the AC generator, with voltage V=V_{0}cos(\omega t) , as shown in the figure. The differential equation involving in this circuit is:

V_{0}cos(\omega t)= RI+L\frac{dI}{dT}+ \frac{1}{C} \int ^t{I dt}
Figure: LCR circuit

Note: Voltage and current in the above equations are time-dependent.

Solving the differential equations is a game of guessing. In the above equation, you have to find the value for I, such that when it is integrated, differentiated, and left alone, the answer should be in the form of V_{0}cos(\omega t) . On integrating and differentiating, the Sine function yields cosine function, but if nothing is done to it remains the same. You see the problem, right? This equation is difficult to solve by guessing.

Rather than solving that equation V_{0} cos(\omega t) , it is easier to solve by slightly modifying the problem. Instead of solving for excitation, it is easier to solve for complex excitation. The differential equation is:

V= RI+L\frac{dI}{dT}+ \frac{1}{C} \int ^t{I dt}

In this equation V and I are complex. L, C and R are real. This is the property of complex numbers. If two complex numbers are equal, then their complex conjugate must be equal. This is an obvious thing. If we apply this property, the differential equation becomes:

V^{*}= RI^{*}+L\frac{dI^{*}}{dT}+ \frac{1}{C} \int ^t{II^{*}dt}

Adding these two equations yields:

(V+V^{*})= R(I+I^{*})+L\frac{d(I+I^{*})}{dT}+ \frac{1}{C} \int ^t{(I+I^{*})dt}

  (V+V^{*}) is 2.Real(V) and (I+I^{*}) is 2.Real(I) . So, the differential equation becomes:

Real(V)= R.Real(I)+L\frac{dReal(I)}{dT}+ \frac{1}{C} \int ^t{Real(I) dt}

This equation tells us that if we solve the problem for complex excitation and get a current value. Then the current value in the original problem, where excitation is a real part of the complex excitation, is a real part of the complex current we get. This actually, is the principle of superposition. The imaginary part of voltage drives the imaginary part of the current and the real part of voltage drives the real part of the current.

Solving the Differential Equation

The advantage of using complex excitation is that if you differentiate it, integrate it, or if you do nothing it remains the same. The differential equation involving complex excitation is:

V_{0} e^{j \omega t} = RI'+L\frac{dI'}{dT}+ \frac{1}{C} \int ^t{I' dt}....................... (I)

The answer to our problem is the real part of the answer to this modified problem. And the solution to this problem is some constant times the complex exponent.

I'=I'_{0} e^{j \omega t}....................... (II)

Using (II) in (I) We get

V_{0} e^{j \omega t} = (j \omega L+ RI'_{0}+\frac{1}{\omega C}) I'_{0} e^{j \omega t}

 

V_{0}= (j \omega L+ RI'_{0}+\frac{1}{\omega C}) I'_{0}

  R+ j(\omega L - \frac{1}{\omega C}) is called the impedance Z of the LCR circuit. Impedance is a complex number with real part equal to   and imaginary part equal to \omega L - \frac{1}{\omega C} . If \phi is the angle made by Z with R ,  we can write:

I'_{0}=\frac{V_{0}}{\mid Z \mid e^{j  \phi}}

 

I'=\frac{V_{0}}{\mid Z \mid } e^{j ( \omega t - \phi)}

But the actual current is the real part of the current that we have calculated. And, we know the real part of the complex number has cosine in it. So, we can write real current as:

\frac{V_{0}}{\mid Z \mid} cos(\omega t - \phi)

This equation tells us that current has magnitude \frac{V_{0}}{\mid Z \mid} and lags the excitation voltage by phase \phi . The interesting thing is, we divide cos (\omega t) by something and we get cos (\omega t - \phi). Think why does this happen?

A Confusion

Some of you may argue with me saying ” There is no such AC generator that produces complex potential.” There is no doubt, such a generator is not possible. Here we use mathematics as a tool. That’s what engineers are famous for, they change the problem, such that it would become easier for solving.

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