# Solving RLC circuit using differential equations

• Engineering
• eehelp150
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eehelp150

## Homework Statement

Find the full response. Assume Vin is a squarewave with Vpp =10V and Vamp = +5V

KCL

## The Attempt at a Solution

My teacher gave this solution but I don't really understand some parts of it.

Full response = Natural response + forced response

Thevenin equivalent:

Rth = R1||R2
Vth = Vin * R2/(R2+R1)

Why is the Thevenin equivalent like this? Is it because the inductor acts like a short and the capacitor acts like an open circuit?

At node 1:
V1 = VC
Why is V1 = VC?
$$\frac{V_C-V_{th}}{R_{th}}+\frac{1}{L}*\int_{0}^{t}V_C (t)+C*\dot{V_C}$$
Move Vth/Rth to other side
$$\frac{V_C}{R_{th}}+\frac{1}{L}*\int_{0}^{t}V_C (t)+C*\dot{V_C}=\frac{V_{th}}{R_{th}}$$
Take the derivative of every part to get rid of integral
$$\dot{\frac{V_C}{R_{th}}}+\dot{(\frac{1}{L}*\int_{0}^{t}V_C(t))} +\dot{C*\dot{V_C}}=\dot{\frac{V_{th}}{R_{th}}}$$
Simplify
$$\dot{\frac{V_C}{R_{th}}}+{\frac{V_C}{L}}+C*\ddot{V_C}=\dot{\frac{V_{th}}{R_{th}}}$$
$$\ddot{V_C}+\dot{\frac{V_C}{R_{th}C}}+\frac{V_C}{LC}=\dot{\frac{V_th}{R_{th}C}}$$

Natural response: $$\ddot{V_N}+\dot{\frac{V_N}{R_{th}C}}+\frac{V_N}{LC}=0$$

Forced response: $$V_F=0$$
Why is Forced response 0?
What is the significance of Vin being a squarewave? How will this affect the final answer?
Would the full response to this problem simply be the Natural response, as Vf = 0?

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eehelp150 said:
Why is the Thevenin equivalent like this? Is it because the inductor acts like a short and the capacitor acts like an open circuit?
Inductor and capacitor are together considered as a load here. So, the Thevenin equivalent will be the equivalent of two resistors and the voltage source.
eehelp150 said:
Why is V1 = VC?
VC−VthRth+​
Because V1 is the potential at node A.
Why have you labelled each element twice? (Source voltage is V1 and V, capacitance is C1 and C..)
In this expression, V1 is the potential of node 1, assuming the -ve terminal of the source to be at ground potential (reference node).
What is the frequency of the source?

cnh1995 said:
Why have you labelled each element twice? (Source voltage is V1 and V, capacitance is C1 and C..)
I think C1 is the name of the device and C is the capacitance, etc.

And http://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ is usually applied to DC circuits. Apparently (if the Thevenin equivalent was provided by teacher) L1 and C1 are considered the load for that circuit ?

BvU said:
And http://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ is usually applied to DC circuits.
It works for any signal, including ac and dc. But calculations for ac signal are easy because of the specially derived phasor algebra for sinusoidal waveform.
Square wave is a sum of infinite sinusoids, so I think phasor algebra doesn't work.

I think the time period of the square wave is greater than the settling time of this circuit and hence, frequency is not mentioned. The circuit is assumed to be in steady state at the end of each half cycle, thus making the forced response i.e.Vc=0.

cnh1995 said:
Inductor and capacitor are together considered as a load here. So, the Thevenin equivalent will be the equivalent of two resistors and the voltage source.
What would be the "process" of "thevenizing" this circuit?
cnh1995 said:
Because V1 is the potential at node A.
Why is it Vc and not Vl or Vr?

cnh1995 said:
Why have you labelled each element twice? (Source voltage is V1 and V, capacitance is C1 and C..)
V1 is the name, V is just something LTspice adds

cnh1995 said:
In this expression, V1 is the potential of node 1, assuming the -ve terminal of the source to be at ground potential (reference node).
What is the frequency of the source?
We are to calculate the frequency of the source so that the capacitor charges for 7tau and discharges for 7tau.

eehelp150 said:
What would be the "process" of "thevenizing" this circuit?
Remove the load i.e. L and C and calculate the volatge across the parallel resistor R. That would be the Thevenin's voltage. Thevenin resistance is the equivalent of the two resistors after "shorting" the voltage source.
eehelp150 said:
Why is it Vc and not Vl or Vr?
Vc=Vr=Vl but here, the forced response is the voltage across the capacitor and all the other quantities are represented in terms of Vc. You can take it as Vl in this case, but be consistent throughout.
eehelp150 said:
We are to calculate the frequency of the source so that the capacitor charges for 7tau and discharges for 7tau.
What's Tau? Is it the time period of the input voltage?

cnh1995 said:
What's Tau? Is it the time period of the input voltage?
Tau is the time constant of the circuit. My professor that once we find the forced response, we need to solve for the characteristic roots. The smallest characteristic root is the time constant. Afterwards:
$$\frac{T}{2}=7\tau\rightarrow T=14\tau\rightarrow f=\frac{1}{T}\rightarrow f=\frac{1}{14\tau} or \frac{characteristic root}{14}$$

## 1. What is an RLC circuit?

An RLC circuit is an electrical circuit that consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series or parallel. These elements can be used to control the flow of current and voltage in a circuit.

## 2. How do you solve an RLC circuit using differential equations?

To solve an RLC circuit using differential equations, we first need to write the differential equation that describes the circuit. This can be done using Kirchhoff's laws and Ohm's law. Then, we can use techniques such as Laplace transforms or the method of undetermined coefficients to solve the differential equation and find the solution for the circuit.

## 3. What are the applications of solving RLC circuits using differential equations?

Solving RLC circuits using differential equations has many applications in electrical and electronic engineering. It can be used to analyze and design electrical circuits, such as filters, amplifiers, and oscillators. It is also used in the study of electromagnetic fields and the behavior of electrical systems.

## 4. What are the challenges of solving RLC circuits using differential equations?

One of the main challenges of solving RLC circuits using differential equations is that the equations can become complex and difficult to solve, especially for circuits with multiple components. Additionally, the initial conditions and boundary conditions of the circuit must be known in order to find a unique solution.

## 5. Can RLC circuits be solved using other methods besides differential equations?

Yes, there are other methods for solving RLC circuits, such as using circuit analysis techniques like Kirchhoff's laws, mesh analysis, and nodal analysis. However, differential equations are often used for more complex circuits or when a more accurate solution is needed.

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