Question

A circuit has in series an electromotive force given by \(E(t)=100 \sin 200 t \mathrm{~V}\), a resistor of \(40 \Omega\), an inductor of \(0.25 \mathrm{H}\), and a capacitor of \(4 \times 10^{-4}\) farads. If the initial current is zero, and the initial charge on the capacitor is \(0.01\) coulombs, find the current at any time \(t>0\).

Short answer

The current in the circuit at any time \(t > 0\) is given by: \(i(t) = A\sin{200t} + B\cos{200t} + C\mathrm{e}^{-rt}\) where A, B, and C are constants determined using the initial conditions and the expressions for the solution derived from the inverse Laplace Transform, and r depends on the components' values in the circuit.

Step-by-step solution

Write the equation for Voltage source, capacitor, inductor, and resistor

Using the provided information, we have the following components in the circuit: - Voltage source: \(E(t) = 100\sin{200t} \mathrm{V}\) - Resistor: \(R = 40\Omega\) - Inductor: \(L = 0.25 \mathrm{H}\) - Capacitor: \(C = 4 \times 10^{-4} \mathrm{F}\) From these components, we can find the voltage drops associated with each component: - Voltage across the resistor: \(V_R(t) = Ri(t)\) - Voltage across the inductor: \(V_L(t) = L\frac{di(t)}{dt}\) - Voltage across the capacitor: \(V_C(t) = \frac{1}{C}\int_{0}^{t}i(t')dt'\)

Apply Kirchhoff's law to create a differential equation

Kirchhoff's voltage law states that the sum of the voltage differences around a loop must be zero. For the given circuit, we have: \(E(t) - V_R(t) - V_L(t) - V_C(t) = 0\) Plugging in the expressions for the voltage drops and rearranging, we get: \(100\sin{200t} = 40i(t) + 0.25\frac{di(t)}{dt} + \frac{1}{4 \times 10^{-4}}\int_{0}^{t}i(t')dt'\)

Transform the equation using Laplace Transforms

Applying the Laplace Transform to both sides of the equation, we get: \(\mathcal{L}\{100\sin{200t}\} = 40\mathcal{L}\{i(t)\} + 0.25\mathcal{L}\{\frac{di(t)}{dt}\} + \frac{1}{4 \times 10^{-4}}\mathcal{L}\{\int_{0}^{t}i(t')dt'\}\) Using the properties of the Laplace Transform and recalling the initial conditions as \(i(0)=0\), and \(Q(0)=0.01C\), we obtain: \(\frac{100}{s^2 + 200^2} = 40I(s) + 0.25sI(s) - 0.25\cdot0 + \frac{1}{4 \times 10^{-4}}\frac{1}{s}I(s)- 0.01s\)

Solve for \(I(s)\) in terms of Laplace transform

Solve the above equation for \(I(s)\): \(I(s) = \frac{(0.25s^2 + 100) + 0.01s^2}{s(40 + 0.25s + \frac{1}{4 \times 10^{-4}})}\)

Use Inverse Laplace Transform Theorem to find \(i(t)\)

Apply the Inverse Laplace Transform to both sides of the last equation: \(i(t) = \mathcal{L}^{-1}\{I(s)\}\) Performing the inverse Laplace transform, the general form of the solution is: \(i(t) = A\sin{200t} + B\cos{200t} + C\mathrm{e}^{-rt}\) Where A, B, and C are constants and r is determined by the denominator of the Laplace transform expression. To determine these constants, we use the initial conditions for current and charge, in addition to the expressions for the solution derived by taking inverse Laplace Transform. Regarding the complexity of calculation to find these constants, you can either proceed algebraically or resort to numerical methods (like partial fraction decomposition). After calculating these constants, you will have the current as a function of time, \(i(t)\), for the given circuit.

Key concepts

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) is a fundamental principle in electrical circuit analysis. It states that the sum of all electrical voltages around any closed loop in a circuit must equal zero. This is based on the law of conservation of energy and applies to both direct current (DC) and alternating current (AC) circuits.

KVL is particularly useful in formulating equations for circuit analysis. By following the path of a loop and summing the voltages produced by batteries, capacitors, resistors, and inductors, one can set up an equation where the net voltage is zero. In our example, this principle is applied to equate the electromotive force (EMF) provided by the sinusoidal voltage source to the sum of the voltage drops across the resistor, inductor, and capacitor.

Applying KVL in Circuit Analysis

When faced with the task of analyzing an electrical circuit, KVL assists in the creation of an ordinary differential equation that models the behavior of the circuit over time. The relationship between the components can be easily expressed, simplifying the process of finding the electrical current at any moment. Students should pay careful attention to the signs of the voltage drops and sources to correctly apply KVL, as misinterpreting these can lead to incorrect equations.

Laplace Transform

The Laplace Transform is a powerful mathematical tool used extensively in engineering, particularly when dealing with differential equations in the time domain. It converts functions of time into functions of a complex variable, making it possible to solve complex problems with relative ease.

One of the Laplace Transform's primary uses in electrical circuit analysis is the simplification of circuits with dynamic components such as capacitors and inductors. It transforms the differential equation derived from Kirchhoff's Law into an algebraic equation in the Laplace domain, where initial conditions are easily taken into account as initial values.

Practical Use of Laplace Transforms in Circuits

Applying the Laplace Transform to our initial value problem allows us to solve for the current, denoted as 'I(s)', in the domain of complex frequencies. This method bypasses the need to work directly with the calculus operations of differentiation and integration in the time domain, which can be quite challenging, especially in circuits with multiple energy storage elements. After using Laplace Transforms, one can employ the Inverse Laplace Transform to translate the solution back to the time domain, revealing the behavior of the circuit over time.

Electrical Circuits Analysis

Electrical Circuits Analysis is the process of examining the components and pathways in an electrical circuit to understand the flow of electric current and the distribution of voltages and resistances throughout the circuit. It's an essential part of designing, testing, and troubleshooting electronic equipment.

In the context of our example, circuit analysis involves identifying the values of resistance, inductance, and capacitance; formulating equations using Kirchhoff's Law; and applying the Laplace Transform to move from the time domain into a more manageable algebraic domain. It also involves considering the initial conditions of the system, such as the initial current and charge on the capacitor, to accurately model and predict the system's behavior.

Key Techniques in Circuit Analysis

Techniques like mesh analysis, nodal analysis, and the use of Thevenin's and Norton's theorems can greatly streamline the process of analyzing complex circuits. The analytical process is iterative and often requires a combination of theoretical knowledge and practical experience to effectively solve for unknown quantities.

Initial Value Problem

An Initial Value Problem in the context of differential equations is a problem where the solution to a differential equation is sought, subject to specific conditions at the start, known as initial conditions. In electrical engineering, this often aligns with the initial current or charge in a circuit component at time zero.

For successful analysis, these initial conditions must be incorporated into the equation solving process, which is elegantly handled by the Laplace Transform in our scenario. Initial conditions are crucial as they can significantly impact the circuit's behavior and the resulting time-domain solution for the current, 'i(t)'.

Crucial Role of Initial Conditions

After transforming the equations with the Laplace Transform and considering the initial conditions, the process involves solving for the transformed function and then finding the inverse to return to the original domain. The initial conditions guide the Inverse Laplace Transform to yield a concrete solution that models the circuit's current at any point in time, precisely describing how the system evolves from its starting point.