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Chapter 8: Problem 37

RC circuit equation Suppose a battery with voltage \(V\) is connected in series to a capacitor (a charge storage device) with capacitance \(C\) and a resistor with resistance \(R\). As the charge \(Q\) in the capacitor increases, the current \(I\) across the capacitor decreases according to the following initial value problems. Solve each initial value problem and interpret the solution. a. \(I^{\prime}(t)+\frac{1}{R C} I(t)=0, I(0)=\frac{V}{R}\) b. \(Q^{\prime}(t)+\frac{1}{R C} Q(t)=\frac{V}{R}, Q(0)=0\)

Short Answer

Expert verified
Question: A capacitor in an electric circuit has an initial charge Q(0) = 0. The charging current is given by the equation \(I'(t) + \frac{1}{RC}I(t) = 0\), where I(t) is the current at time t, R is the resistance, C is the capacitance, and V is the voltage of the battery. Find the charge on the capacitor, Q(t), after time t. Answer: The charge on the capacitor, Q(t), after time t is given by the equation \(Q(t) = V R \left(1 - e^{-\frac{t}{RC}}\right)\).

Step by step solution

01

a. Solve: \(I'(t) + \frac{1}{RC}I(t) = 0, I(0) = \frac{V}{R}\)

First, find the general solution for \(I(t)\) by solving the given differential equation: \(I'(t) + \frac{1}{RC}I(t) = 0\). This is a linear, first-order, homogeneous differential equation, and it can be solved using an integrating factor. The integrating factor is given by \(\mu(t)=e^{\int\frac{1}{RC}\,dt}=e^{\frac{t}{RC}}\). Now, multiply the given equation by the integrating factor \(\mu (t)\): \(e^{\frac{t}{RC}}I'(t) + \frac{1}{RC}e^{\frac{t}{RC}}I(t) = 0\) Now, we can notice that the left side of the equation is the derivative of a product \(I(t)e^{\frac{t}{RC}}\). So integrate both sides with respect to \(t\): \(\int \left( I'(t)e^{\frac{t}{RC}} + \frac{1}{RC}e^{\frac{t}{RC}} I(t) \right) dt = \int 0 \, dt\) \(I(t)e^{\frac{t}{RC}} = C\) where \(C\) is an integration constant. To find the general solution for \(I(t)\), solve for it by dividing both sides by \(e^{\frac{t}{RC}}\): \(I(t)=Ce^{-\frac{t}{RC}}\) Now use the initial condition \(I(0) = \frac{V}{R}\) to find the particular solution: \(I(0)=Ce^{-\frac{0}{RC}}=C=C_1=\frac{V}{R}\) Thus, the particular solution for this problem is: \(I(t)=\frac{V}{R}e^{-\frac{t}{RC}}\)
02

b. Solve: \(Q'(t) + \frac{1}{RC}Q(t) = \frac{V}{R}, Q(0) = 0\)

This time, we need to find the general solution for \(Q(t)\) by solving the given differential equation: \(Q'(t) + \frac{1}{RC}Q(t) = \frac{V}{R}\). This is a linear, first-order, non-homogeneous differential equation. We will also use an integrating factor to solve it. The integrating factor is the same as in part a since the coefficient of \(Q(t)\) is the same: \(\mu(t)=e^{\int\frac{1}{RC}\,dt}=e^{\frac{t}{RC}}\). Now, multiply the given equation by the integrating factor \(\mu (t)\): \(e^{\frac{t}{RC}}Q'(t) + \frac{1}{RC}e^{\frac{t}{RC}}Q(t) = \frac{V}{R}e^{\frac{t}{RC}}\) Once again, notice that the left side of the equation is the derivative of a product \(Q(t)e^{\frac{t}{RC}}\). So integrate both sides with respect to \(t\): \(\int \left( Q'(t)e^{\frac{t}{RC}} + \frac{1}{RC}e^{\frac{t}{RC}} Q(t) \right) dt = \int \frac{V}{R} e^{\frac{t}{RC}}\, dt\) \(Q(t)e^{\frac{t}{RC}} = \frac{V}{R} \int e^{\frac{t}{RC}}\, dt\) \(Q(t)e^{\frac{t}{RC}} = \frac{V R}{R} e^{\frac{t}{RC}} + C_2\) Now, solve for \(Q(t)\) by dividing both sides by \(e^{\frac{t}{RC}}\): \(Q(t)=V R e^{-\frac{t}{RC}} + C_2 e^{-\frac{t}{RC}}\) To find the particular solution, use the initial condition \(Q(0) = 0\): \(Q(0) = V R e^{-\frac{0}{RC}} + C_2 e^{-\frac{0}{RC}} = VR + C_2\) It follows that \(C_2=-VR\), so the particular solution for this problem is: \(Q(t)= V R \left(1 - e^{-\frac{t}{RC}}\right)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations in RC Circuits
Understanding the behavior of electrical circuits often involves solving differential equations. An RC circuit, which includes a resistor (R) and a capacitor (C), is a great example. These circuits can be described using linear first-order differential equations.

For example, the equation \(I'(t) + \frac{1}{RC}I(t) = 0\) describes the change in current \(I(t)\) over time. This type of equation is homogeneous, meaning there is no external force influencing it.

Solving it involves finding a function that satisfies this equation, often using techniques like the integrating factor method. This provides insights into how the current in the circuit behaves as the capacitor charges or discharges.
Capacitor Charging Dynamics
When a capacitor charges in an RC circuit, its behavior can be modeled by a differential equation. As the capacitor stores charge, the current flowing through it changes. The equation \(Q'(t) + \frac{1}{RC}Q(t) = \frac{V}{R}\) illustrates this process, where \(Q(t)\) is the charge on the capacitor and the source voltage is \(V\).

This is a non-homogeneous equation, indicating that the system is influenced by an external voltage source. The solution describes how the charge builds up over time, initially increasing rapidly and then leveling off as it approaches its maximum capacity.

Understanding this charging process helps in designing circuits with desired time-dependent behaviors for applications like timing and filtering.
Initial Value Problem
When dealing with RC circuits, an important aspect is solving the initial value problem. This involves using initial conditions to find a specific solution to the differential equation that describes the circuit.

For instance, the initial condition \(I(0) = \frac{V}{R}\) ensures that the solution reflects the circuit's behavior precisely at the moment the circuit starts operating.
  • It provides a starting point to determine how the current or charge changes over time.
  • It helps ensure accuracy in modeling real-life circuits.

By incorporating initial values, students can predict the exact behavior of electrical components in a circuit, enhancing their understanding of circuit dynamics.

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Most popular questions from this chapter

A fish hatchery has 500 fish at \(t=0\), when harvesting begins at a rate of \(b>0\) fish/year. The fish population is modeled by the initial value problem \(y^{\prime}(t)=0.01 y-b, y(0)=500,\) where \(t\) is measured in years. a. Find the fish population, for \(t \geq 0\), in terms of the harvesting rate \(b\) b. Graph the solution in the case that \(b=40\) fish/year. Describe the solution. c. Graph the solution in the case that \(b=60\) fish/year. Describe the solution.

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\frac{e^{t}}{2 y}, y(\ln 2)=1$$

Determine whether the following equations are separable. If so, solve the initial value problem. $$\sec x y^{\prime}(x)=y^{3}, y(0)=3$$

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t).$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{2 t}{t^{2}+1} y(t)=1+3 t^{2}, \quad y(1)=4$$

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$t^{3} y^{\prime}(t)+3 t^{2} y=\frac{1+t}{t}, y(1)=6$$
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