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Chapter 5: Problem 3

A circuit has in series a constant electromotive force of \(100 \mathrm{~V}\), a resistor of \(10 \Omega\), and a capacitor of \(2 \times 10^{-4}\) farads. The switch is closed at time \(t=0\), and the charge on the capacitor at this instant is zero. Find the charge and the current at time \(t>0\),

Short Answer

Expert verified
The charge on the capacitor at any time \(t > 0\) can be found using the expression \(Q(t) = 100(1 - e^{-\frac{t}{(10)(2 \times 10^{-4})}})C\), while the current at any time \(t > 0\) can be found using the expression \(I(t) = 10e^{-\frac{t}{(10)(2 \times 10^{-4})}}A\).

Step by step solution

01

1. Kirchhoff's Loop Rule

Let's start by applying Kirchhoff's loop rule to our circuit. The loop rule states that the sum of voltages across all components in a loop is equal to zero. In our loop, we have the electromotive force E, the resistor with resistance R, and the capacitor with capacitance C. Therefore, the loop rule equation becomes: \[ E - I(t)R - V_C(t) = 0 \] where \(I(t)\) is the current at time t, \(V_C(t)\) is the voltage across the capacitor at time t.
02

2. Calculate the voltage across the capacitor

We need to find the voltage across the capacitor as a function of time. The voltage across a capacitor is given by \[V_C(t) = Q(t) / C\] where Q(t) is the charge on the capacitor at time t. Therefore, the loop rule equation becomes: \[ E - I(t)R - \frac{Q(t)}{C} = 0 \]
03

3. Rewrite the loop rule equation in terms of current and charge

Next, we need to express the current in terms of the charge. Since the current is the rate of change of charge with respect to time, we have: \[I(t) = \frac{dQ(t)}{dt}\] We can plug this into our loop rule equation to get: \[ E - R\frac{dQ(t)}{dt} - \frac{Q(t)}{C} = 0 \]
04

4. Solve the differential equation for the charge

The equation we derived in Step 3 is a first-order linear ordinary differential equation. We can rewrite it as: \[\frac{dQ(t)}{dt} + \frac{Q(t)}{RC} = \frac{E}{R}\] Now we can use integration factor method or other methods to solve this differential equation. The solution is: \[Q(t) = CE(1 - e^{-\frac{t}{RC}})\]
05

5. Calculate the current as a function of time

Now that we have the charge on the capacitor as a function of time, we can calculate the current by simply taking the derivative of our charge equation: \[I(t) = \frac{dQ(t)}{dt} = \frac{E}{R}e^{-\frac{t}{RC}}\] Now we have the charge and the current as functions of time and we can substitute the given values to find the charge and the current at any time t > 0: For our given values: \(E = 100V\), \(R = 10\Omega\), and \(C = 2 \times 10^{-4}F\), thus: \[Q(t) = 100(1 - e^{-\frac{t}{(10)(2 \times 10^{-4})}})C\] \[I(t) = \frac{100}{10}e^{-\frac{t}{(10)(2 \times 10^{-4})}}A\] These expressions give the charge on the capacitor and the current, respectively, at any time t > 0.

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Key Concepts

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

Kirchhoff's Loop Rule
Kirchhoff's Loop Rule is a fundamental principle in electrical circuit theory. This rule, part of Kirchhoff's circuit laws, states that the sum of the potential differences (voltages) around any closed loop in a circuit must equal zero. In practical terms, it means that whatever energy is supplied to the circuit, usually by a battery or other sources, must be fully accounted for in energy drops across components like resistors and capacitors. When using this rule, keep in mind to follow the passive sign convention: traversing a component in the same direction as the current yields a drop in potential, whereas traversing against yields a rise.
In our example of a series RC circuit, the constant electromotive force (EMF) provides energy that is then dissipated by the resistor and stored in the capacitor. This energy conservation is mathematically expressed through the loop rule equation which includes the voltage supplied by the EMF, the current through the resistor (defined by Ohm's Law), and the voltage across the capacitor, thus creating a relationship that is essential to solving for the unknown values of charge and current at a given time.
Capacitor Charge and Current
Understanding the relationship between capacitor charge and current is crucial in analyzing RC circuits. A capacitor stores energy in the form of an electric field, which is established by the separation of charges on its plates. To understand how a capacitor interacts with the circuit, it's important to know that the voltage across a capacitor is directly proportional to the charge it holds, as given by the equation \(V_C(t) = Q(t) / C\).
Additionally, the current in the circuit is linked to how this charge changes over time; mathematically represented by the relation \(I(t) = \frac{dQ(t)}{dt}\). Here, current is seen as the rate of charge flow, where a positive current denotes charge flowing into the capacitor's positive plate. This understanding leads to the insight needed for establishing the loop rule equation in terms of charge and current and is indispensable for solving the circuit's differential equation.
Solving Ordinary Differential Equations
Solving ordinary differential equations (ODEs) is an important aspect in modeling the behavior of circuits. An ODE involves functions of one variable and their derivatives, pertinent to dynamic systems. In contexts such as our RC circuit problem, an ODE describes how the charge on the capacitor evolves over time.
To solve an ODE, various methods are available, including separation of variables, integrating factors, and characteristic equations. In our case, an integrating factor is a common technique used to solve linear first-order ODEs, such as the one derived from the loop rule. The solution to this ODE gives us a function that represents the charge on the capacitor as a function of time which, in turn, allows us to express and compute the current. This process demystifies how the charge and current change as the capacitor charges and plays a crucial role in understanding how circuits operate over time.
Exponential Decay in Circuits
Exponential decay is a ubiquitous concept in circuit analysis, especially when dealing with RC circuits like the one in our exercise. When a capacitor in such a circuit is charging or discharging through a resistor, the charge and current exhibit an exponential behavior over time. This phenomenon is characterized by the capacitor's ability to store and release energy and is mathematically expressed by the exponential term in the solution for capacitor charge \(Q(t) = CE(1 - e^{-\frac{t}{RC}})\) and current \(I(t) = \frac{E}{R}e^{-\frac{t}{RC}}\).
In these expressions, the term \(RC\) is known as the time constant of the circuit and symbolizes the speed at which the charging and discharging processes occur. For students and engineers alike, visualizing the exponential graph of charge and current as they change over time provides valuable intuition into the transient behavior of the circuit and the time scale over which it operates.

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Most popular questions from this chapter

A 4-lb weight is hung upon the lower end of a coil spring hanging vertically from a fixed support. The weight comes to rest in its equilibrium position, thereby stretching the spring 8 in. The weight is then pulled down a certain distance below this equilibrium position and released at \(t=0\). The medium offers a resistance in pounds numerically equal to \(a(d x / d t)\), where \(a>0\) and \(d x / d t\) is the instantaneous velocity in feet per second. Show that the motion is oscillatory if \(a<\sqrt{3}\), critically damped if \(a=\sqrt{3}\), and overdamped if \(a>\sqrt{3}\).

A 16-lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring \(0.4 \mathrm{ft}\). Then, beginning at \(t=0\), an external force given by \(F(t)=40 \cos 16 t\) is applied to the system. The medium offers a resistance in pounds numerically equal to \(4(d x / d t)\), where \(d x / d t\) is the instantaneous velocity in feet per second. (a) Find the displacement of the weight as a function of the time. (b) Graph separately the transient and steady-state terms of the motion found in step (a) and then use the curves so obtained to graph the entire displacement itself.

A 4 -lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 in. At time \(t=0\) the weight is then struck so as to set it into motion with an initial velocity of \(2 \mathrm{ft} / \mathrm{sec}\), directed downward. (a) Determine the resulting displacement and velocity of the weight as functions of the time. (b) Find the amplitude, period, and frequency of the motion. (c) Determine the times at which the weight is \(1.5\) in. below its equilibrium position and moving downward. (d) Determine the times at which it is \(1.5\) in. below its equilibrium position and moving upward.

A simple pendulum is composed of a mass \(m\) (the bob) at the end of a straight wire of negligible mass and length \(L\). It is suspended from a fixed point \(S\) (its point of support) and is free to vibrate in a vertical plane (sec Figure 5.4). Let SP denote the straight wire; and let \(\theta\) denote the angle that \(S P\) makes with the vertical \(S P_{0}\) at time \(t\), positive when measured counterclock wise. We neglect air resistance and assume that only two forces act on the mass \(m: F_{1}\), the tension in the wire; and \(F_{2}\), the force due to gravity, which acts vertically downward and is of magnitude mg. We write \(F_{2}=F_{r}+F_{N}\), where \(F_{r}\) is the component of \(F_{2}\) along the tangent to the path of \(m\) and \(F_{N}\) is the component of \(F_{2}\) normal to \(F_{r}\). Then \(F_{N}=-F_{1}\) and \(F_{\mathrm{r}}=-m g \sin \theta\), and so the net force acting on \(m\) is \(F_{1}+F_{2}=F_{1}+F_{r}+F_{N}=-m g \sin \theta\), along the arc \(P_{0} P\). Letting s denote the length of the arc \(P_{0} P\), the acceleration along this arc is \(d^{2} s / d t^{2} .\) Hence applying Newton's second law, we have \(m d^{2} s / d t^{2}=-m g \sin \theta\) But since \(s=1 \theta\), this reduces to the differential equation $$ m l \frac{d^{2} \theta}{d t^{2}}=-m g \sin \theta \quad \text { or } \quad \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0 . $$ (a) The equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0 $$ is a nonlinear second-order differential equation. Now recall that $$ \sin \theta=\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}-\cdots \cdot $$ Hence if \(\theta\) is sufficiently small, we may replace \(\sin \theta\) by \(\theta\) and consider the approximate linear equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \theta=0 $$ Assume that \(\theta=\theta_{0}\) and \(d \theta / d t=0\) when \(t=0\). Obtain the solution of this approximate equation that satisfies these initial conditions and find the amplitude and period of the resulting solution. Observe that this period is independent of the initial displacement. (b) Now return to the nonlinear equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0 $$ Multiply through by \(2 d \theta / d t\), integrate, and apply the initial condition \(\theta=\theta_{\text {o. }}\) \(d \theta / d t=0 .\) Then separate variables in the resulting equation to obtain $$ \frac{d \theta}{\sqrt{\cos \theta-\cos \theta_{0}}}=\pm \sqrt{\frac{2 g}{l}} d t . $$ From this equation determine the angular velocity \(d \theta / d t\) as a function of \(\theta\). Note that the left member cannot be integrated in terms of elementary functions to obtain the exact solution \(\theta(t)\) of the nonlinear differential equation.

A 16 -lb weight is attached to the lower end of a coil spring that is suspended vertically from a support and for which the spring constant \(k\) is \(10 \mathrm{lb} / \mathrm{ft}\). The weight comes to rest in its equilibrium position and is then pulled down 6 in. below this position and released at \(t=0 .\) At this instant the support of the spring begins a vertical oscillation such that its distance from its initial position is given by \(\frac{1}{2} \sin 2 t\) for \(t \geq 0\). The resistance of the medium in pounds is numerically equal to \(2(d x / d t)\), where \(d x / d t\) is the instantaneous velocity of the moving weight in feet per second. (a) Show that the differential equation for the displacement of the weight from its equilibrium position is $$ \frac{1}{2} \frac{d^{2} x}{d t^{2}}=-10(x-y)-2 \frac{d x}{d t}, \quad \text { where } \quad y=\frac{1}{2} \sin 2 t $$ and hence that this differential equation may be written $$ \frac{d^{2} x}{d t^{2}}+4 \frac{d x}{d t}+20 x=10 \sin 2 t $$ (b) Solve the differential equation obtained in step (a), apply the relevant initial conditions, and thus obtain the displacement \(x\) as a function of time.
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