Professor Kate Poirier | OL67 | Fall 2020

# Category: Project #3 groups

Spring Problems I focuses on undamped spring systems. That is, a spring system that lacks any forces counteracting the restorative force of the spring itself. This is a purely theoretical exercise as there are usually energy loses due to heat, drag against a resistive medium like water or air, a dedicated dampening force, or other external forces.

Fig 1 shows three key positions when discussing spring systems. L represents the unloaded spring length. “l” represents the spring elongation, sometimes using “s” for notation. When there is a mass on the spring and the spring system is at rest it is said to be at equilibrium. “y” represents an additional displacement beyond the equilibrium point, in this case either above or below.

Partner Group: Problem 1:

Partner Group: Problem 3:

Partner Group: Problem 4:

Partner Group: Problem 11:

Partner Group: Problem 13:

Partner Group: Problem 19:

Partner Group: Problem 21:

6.3 The RLC Circuit

Introduction:

The main purpose of chapter 6.3 is to show how differential equations can be used to solve RLC Circuits problems. RLC circuits can simply be explained as electrical circuits that consist of a resistor, an inductor, and a capacitor all connected to each other. Now to understand the real world problem part of this chapter, the function of the components must be understood. The resistor has many uses, in an electrical circuit one of its main purposes is to reduce current flow. The inductor protects the circuit from changes in the current that can damage other components. Finally the capacitor stores electric charges that can be denoted as (Q) which can be used for other purposes. In other words differences in potential occur at the resistor, induction coil, and the capacitor. RLC circuits have many applications that range from oscillator circuits to radio receivers.

Now in chapter 6.3 we are mainly concerned with the difference in electric potential that occurs in a closed circuit, which is caused by the flow of a current in the circuit. In other words we are interested in the changes that occur in a RLC circuit when voltage(E) is introduced to the circuit producing a current(I) that runs across the circuit, and all the other components (R), (L), (C) finally giving us the electrical potential(E(t)). Now that we know what exactly we want to find out in the real world problem. This section is solving RLC circuit problems through the use of Linear Second Order Equations. In the following part of this lesson will be showing the ways in which all the units relate to each other and what formulas are used to solve the RLC problems. In the image above we have an example of an RLC circuit and all the units need it to solve the RLC circuit problems. The following notations are used:E= E(t)  for the electric potential/impressed voltage from a battery or generator I=I(t) for the current of the circuitR is the resistance of the resistorL is the inductance of the inductorC is the capacitance of the capacitor

In accordance with Kirchhoff’s Law, it is stated that the sum of the voltage drops in a closed RLC circuit equals the impressed voltage. Using formulas 1,2,4  from the section above we get the new formula:

Same as in Second Order Linear Equation problems there are also initial value problems.

In the example section several example problems are able to explain how these equations are solved.

Finally we also have Non-homogeneous Second Linear Equation Problems

This problem deals with forced oscillation with damping, a more thorough explanation is given in example 5 where a Non-homogeneous Second Linear Equation problem is explained and solved with the “undetermined coefficients” method.

Examples of RLC Circuit Problems:

Example 1:

Find the current in the RLC circuit, assuming that E(t) = 0

for t > 0.

R = 3 ohms; L = 0.1 henrys; C = 0.01 farads; Q0 = 0 coulombs; I0 = 2 amperes.

The video example and Solution is here.

Example 2:

Find the current in the RLC circuit, assuming that E(t) = 0

for t > 0.

R = 2 ohms; L = 0.05 henrys; C = 0.01 farads’; Q0 = 2 coulombs; I0 = 2 amperes.

The video example and Solution is here.

Example 3:

Find the current in the RLC circuit, assuming that E(t) = 0

for t > 0.

R = 2 ohms; L = 0.1 henrys; C = 0.01 farads; Q0 = 2 coulombs; I0 = 0 amperes.

Example 4:

Find the steady state current in the circuit describe by the equation.

1/10Q”+6Q’+250Q = 10 cos 100t + 30 sin100t

The video example and solution is here.

Find the steady state current in the circuit describe by the equation.

1/10Q”+3Q’+100Q = 5cos 10t – 5sin 10t

The video example and Solution is here.

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