Problem 34
Question
Suppose an \(R C\) -series circuit has a variable resistor. If the resistance at time \(t\) is given by \(R=k_{1}+k_{2} t\), where \(k_{1}\) and \(k_{2}\) are known positive constants, then (9) becomes $$ \left(k_{1}+k_{2} t\right) \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ If \(E(t)=E_{0}\) and \(q(0)=q_{0}\), where \(E_{0}\) and \(q_{0}\) are constants, show that $$ q(t)=E_{0} C+\left(q_{0}-E_{0} C\right)\left(\frac{k_{1}}{k_{1}+k_{2} t}\right)^{1 / C k_{2}} $$
Step-by-Step Solution
Verified Answer
The solution is \(q(t) = E_0 C + (q_0 - E_0 C)\left(\frac{k_1}{k_1+k_2t}\right)^{1/(Ck_2)}\).
1Step 1: Identify the Differential Equation
The given differential equation is \((k_1 + k_2 t) \frac{dq}{dt} + \frac{1}{C}q = E_0\). This equation represents a first-order linear non-homogeneous ordinary differential equation.
2Step 2: Simplify the Differential Equation
Divide the entire equation by the term \(k_1 + k_2 t\) to put it into the standard linear form: \(\frac{dq}{dt} + \frac{1}{C(k_1 + k_2 t)}q = \frac{E_0}{k_1 + k_2 t}\).
3Step 3: Find the Integrating Factor
To solve the linear DE, find the integrating factor \(\mu(t) = \exp\left(\int \frac{1}{C(k_1 + k_2 t)} dt\right)\). After calculating, \(\mu(t) = \left(k_1 + k_2 t\right)^{1/(Ck_2)}\).
4Step 4: Multiply the Differential Equation by the Integrating Factor
Multiply each term of the simplified equation by \(\mu(t)\) to get: \(\left(k_1 + k_2 t\right)^{1/(Ck_2)} \frac{dq}{dt} + \frac{1}{C} \left(k_1 + k_2 t\right)^{1/(Ck_2) - 1} q = E_0 \left(k_1 + k_2 t\right)^{1/(Ck_2) - 1}\).
5Step 5: Integrate to Find q(t)
Recognize the left side as the derivative of \(\left(k_1 + k_2 t\right)^{1/(Ck_2)} q\) and integrate both sides: \(\int d\left(\left(k_1 + k_2 t\right)^{1/(Ck_2)} q\right) = \int E_0 \left(k_1 + k_2 t\right)^{1/(Ck_2) - 1} dt\).
6Step 6: Solve the Integral
Integrate the right side, which gives \(E_0 C \left(k_1 + k_2 t\right)^{1/(Ck_2)} + C_1\), where \(C_1\) is a constant.
7Step 7: Find the Constant from Initial Condition
Use the initial condition \(q(0) = q_0\) to find \(C_1\). Substitute \(t = 0\) into \(q(0) = E_0 C (k_1)^{1/(Ck_2)} + C_1\) to solve for \(C_1: C_1 = q_0 - E_0 C (k_1)^{1/(Ck_2)}\).
8Step 8: Substitute Back to Find q(t)
Substitute \(C_1\) back into the solution to solve for \(q(t)\): \(q(t) = E_0 C + \left(q_0 - E_0 C\right)\left(\frac{k_1}{k_1 + k_2 t}\right)^{1/(Ck_2)}\).
Key Concepts
RC Circuit AnalysisFirst-order Linear Differential EquationIntegrating Factor MethodElectrical Engineering Mathematics
RC Circuit Analysis
An RC circuit, short for Resistor-Capacitor circuit, is a fundamental concept in electrical engineering. It consists of resistors and capacitors connected in series or parallel. These components store and release energy within the circuit. By analyzing an RC circuit, engineers can predict its behavior over time, such as voltage changes or current flow, which is crucial for designing efficient and functional electronic systems.
A key aspect of RC circuit analysis is understanding how the capacitor charges and discharges through the resistor. This process is governed by a differential equation, which models the interplay between current, voltage, and time in the circuit. Understanding these dynamics helps in designing circuits with desired electrical properties.
A key aspect of RC circuit analysis is understanding how the capacitor charges and discharges through the resistor. This process is governed by a differential equation, which models the interplay between current, voltage, and time in the circuit. Understanding these dynamics helps in designing circuits with desired electrical properties.
- RC circuits are used in a variety of applications, from filtering signals to timing functions.
- The time constant, denoted by the Greek letter \( \tau \) (tau), is a crucial parameter that provides insights into how quickly the circuit responds to external changes.
First-order Linear Differential Equation
First-order linear differential equations are a type of equation that describe how a quantity changes over time. They are 'first-order' because they involve the first derivative \(\frac{dq}{dt}\) and 'linear' because the variable \(q\) and its derivative appear to the first power without being multiplied together.
In the context of an RC circuit with a variable resistor, the equation \((k_1 + k_2 t) \frac{dq}{dt} + \frac{1}{C}q = E_0\) models the changing charge \(q\) on the capacitor over time \(t\). Understanding this allows us to predict how the circuit reacts to a constant external voltage source \(E(t) = E_0\).
In the context of an RC circuit with a variable resistor, the equation \((k_1 + k_2 t) \frac{dq}{dt} + \frac{1}{C}q = E_0\) models the changing charge \(q\) on the capacitor over time \(t\). Understanding this allows us to predict how the circuit reacts to a constant external voltage source \(E(t) = E_0\).
- These equations are crucial for modeling systems where the rate of change depends on both time and the evolving state.
- Solving them helps determine how systems, such as RC circuits, evolve over time given specific conditions.
Integrating Factor Method
The Integrating Factor Method is a powerful technique used to solve first-order linear differential equations. It works by transforming the equation into a form that is easy to integrate. This involves finding an "integrating factor," a function that simplifies the process.
In our RC circuit problem, the integrating factor is chosen based on the coefficient of \(q\) in the simplified differential form. Here, that coefficient is \(\frac{1}{C(k_1 + k_2 t)}\), so the integrating factor \( \mu(t) \) is calculated as \( \mu(t) = \exp\left(\int \frac{1}{C(k_1 + k_2 t)} dt\right) \).This technique helps turn complex differential equations into simpler ones that can be tackled with straightforward integration techniques.
In our RC circuit problem, the integrating factor is chosen based on the coefficient of \(q\) in the simplified differential form. Here, that coefficient is \(\frac{1}{C(k_1 + k_2 t)}\), so the integrating factor \( \mu(t) \) is calculated as \( \mu(t) = \exp\left(\int \frac{1}{C(k_1 + k_2 t)} dt\right) \).This technique helps turn complex differential equations into simpler ones that can be tackled with straightforward integration techniques.
- It converts the differential equation into an exact differential, allowing easy integration.
- Allows for systematic solutions to non-homogeneous differential equations commonly found in engineering problems.
Electrical Engineering Mathematics
Electrical Engineering Mathematics is a field of study that focuses on applying mathematical techniques to solve electrical engineering problems. This includes the use of differential equations, complex numbers, and linear algebra.
In the context of RC circuit analysis, differential equations play a critical role in predicting circuit behavior over time. Understanding these mathematical concepts enables engineers to design circuits that perform specific electrical functions efficiently.
Electrical engineering mathematics is essential for modeling and analyzing circuits and systems, enabling the optimization of various electrical components and systems.
In the context of RC circuit analysis, differential equations play a critical role in predicting circuit behavior over time. Understanding these mathematical concepts enables engineers to design circuits that perform specific electrical functions efficiently.
Electrical engineering mathematics is essential for modeling and analyzing circuits and systems, enabling the optimization of various electrical components and systems.
- Mathematics forms the backbone for the theoretical understanding and practical application in electrical engineering.
- Advanced topics like control theory, signal processing, and communications heavily rely on robust mathematical foundations.
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Problem 33
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