An electric circuit, consisting of a capacitor, resistor, and an electromotive force can be modeled by the differential equation bp dt 1 + R = E(t), where R and C are constants (resistance and capacitance) and = b q(t) is the amount of charge on the capacitor at time t. For simplicity in the following analysis, let R = C = 1, forming the differential equation dq/dt + q = E(t). In Exer- cises 17–20, an electromotive force is given in piecewise form, a favorite among engineers. Assume that the initial charge on the capacitor is zero [q(0) = 0]. (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval [0, 4]. (ii) Hind an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period. 2t, if 0 < t < 2, 19. Е(() — 0, if t > 2
An electric circuit, consisting of a capacitor, resistor, and an electromotive force can be modeled by the differential equation bp dt 1 + R = E(t), where R and C are constants (resistance and capacitance) and = b q(t) is the amount of charge on the capacitor at time t. For simplicity in the following analysis, let R = C = 1, forming the differential equation dq/dt + q = E(t). In Exer- cises 17–20, an electromotive force is given in piecewise form, a favorite among engineers. Assume that the initial charge on the capacitor is zero [q(0) = 0]. (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval [0, 4]. (ii) Hind an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period. 2t, if 0 < t < 2, 19. Е(() — 0, if t > 2
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Please answer 19 part ii showing the steps to arrive at the solution also attached in a pic.
![**Analysis of an Electric Circuit Using Differential Equations**
An electric circuit, which includes a capacitor, resistor, and an electromotive force, can be described using the differential equation:
\[ R \frac{dq}{dt} + \frac{1}{C} q = E(t), \]
where \( R \) and \( C \) are constants (resistance and capacitance, respectively), and \( q = q(t) \) is the charge on the capacitor at time \( t \). For simplicity in our analysis, we assume \( R = C = 1 \), simplifying the differential equation to:
\[ \frac{dq}{dt} + q = E(t). \]
In Exercises 17-20, the electromotive force is given in piecewise form:
\[
E(t) = \begin{cases}
2t, & \text{if } 0 < t < 2, \\
0, & \text{if } t \geq 2
\end{cases}
\]
**Assumptions:**
- The initial charge on the capacitor is zero, i.e., \( q(0) = 0 \).
**Tasks:**
1. (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval \([0, 4]\).
2. (ii) Find an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period.
**Explicit Solution:**
To solve the differential equation \(\frac{dq}{dt} + q = E(t)\) with the given piecewise function \(E(t)\), we separate the analysis into two intervals:
1. For \(0 < t < 2\), \(E(t) = 2t\),
2. For \(t \geq 2\), \(E(t) = 0\).
The solution involves solving for \(q(t)\) in each interval and applying the initial condition \(q(0) = 0\).
Analyzing these intervals:
**Interval 1: \(0 < t < 2\)**
The differential equation is:
\[ \frac{dq}{dt} + q = 2t \]
Applying the method of integrating factors, we solve for \(q(t)\).
**Interval 2: \(t \geq 2\)**
Here, the differential equation simplifies to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b1372ae-e668-4971-89ee-29da7ac7c466%2F021de45a-9dbf-4bc3-b97c-754e55dc5561%2Fz31k65_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Analysis of an Electric Circuit Using Differential Equations**
An electric circuit, which includes a capacitor, resistor, and an electromotive force, can be described using the differential equation:
\[ R \frac{dq}{dt} + \frac{1}{C} q = E(t), \]
where \( R \) and \( C \) are constants (resistance and capacitance, respectively), and \( q = q(t) \) is the charge on the capacitor at time \( t \). For simplicity in our analysis, we assume \( R = C = 1 \), simplifying the differential equation to:
\[ \frac{dq}{dt} + q = E(t). \]
In Exercises 17-20, the electromotive force is given in piecewise form:
\[
E(t) = \begin{cases}
2t, & \text{if } 0 < t < 2, \\
0, & \text{if } t \geq 2
\end{cases}
\]
**Assumptions:**
- The initial charge on the capacitor is zero, i.e., \( q(0) = 0 \).
**Tasks:**
1. (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval \([0, 4]\).
2. (ii) Find an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period.
**Explicit Solution:**
To solve the differential equation \(\frac{dq}{dt} + q = E(t)\) with the given piecewise function \(E(t)\), we separate the analysis into two intervals:
1. For \(0 < t < 2\), \(E(t) = 2t\),
2. For \(t \geq 2\), \(E(t) = 0\).
The solution involves solving for \(q(t)\) in each interval and applying the initial condition \(q(0) = 0\).
Analyzing these intervals:
**Interval 1: \(0 < t < 2\)**
The differential equation is:
\[ \frac{dq}{dt} + q = 2t \]
Applying the method of integrating factors, we solve for \(q(t)\).
**Interval 2: \(t \geq 2\)**
Here, the differential equation simplifies to
![The exact solution is given by the following piecewise function:
\[ q(t) = \begin{cases}
2(t - 1 + e^{-t}), & \text{if } 0 < t < 2, \\
2(1 + e^{-2})e^{2-t}, & \text{if } t \geq 2
\end{cases} \]
To evaluate \( q \) at \( t = 4 \):
\[ q(4) = 2(1 + e^{-2})e^{-2} \approx 0.3073. \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b1372ae-e668-4971-89ee-29da7ac7c466%2F021de45a-9dbf-4bc3-b97c-754e55dc5561%2Fu0k41kq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The exact solution is given by the following piecewise function:
\[ q(t) = \begin{cases}
2(t - 1 + e^{-t}), & \text{if } 0 < t < 2, \\
2(1 + e^{-2})e^{2-t}, & \text{if } t \geq 2
\end{cases} \]
To evaluate \( q \) at \( t = 4 \):
\[ q(4) = 2(1 + e^{-2})e^{-2} \approx 0.3073. \]
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