What Is The Analytic Solution For The Voltage Across The Resistor?

Introduction

When a capacitor is connected to a resistor in a circuit, the voltage across the capacitor changes over time due to the flow of current. In this scenario, we have three capacitors with a capacitance of 1F each, and one of them (C1) has an initial voltage of 5V, while the other two have an initial voltage of 0V. A resistor with a resistance of 1Ω is connected in series with the capacitors. At time t=0, the switch closes, completing the circuit. The question asks for the analytic solution for the voltage across the resistor.

Understanding the Circuit

To solve this problem, we need to understand the circuit and the behavior of the capacitor and resistor. When the switch closes, the capacitor starts to charge, and the voltage across it changes over time. The resistor opposes the flow of current, causing the voltage across the capacitor to increase gradually. We can model this behavior using the differential equation for a capacitor and resistor in series.

Differential Equation for Capacitor and Resistor

The differential equation for a capacitor and resistor in series is given by:

$$\frac{dV}{dt} = -\frac{1}{RC}V$$

where V is the voltage across the capacitor, R is the resistance, C is the capacitance, and t is time.

Initial Conditions

We are given the following initial conditions:

• C1 has an initial voltage of 5V
• C2 and C3 have an initial voltage of 0V
• The switch closes at time t=0

Analytic Solution

To find the analytic solution for the voltage across the resistor, we need to solve the differential equation. We can do this by separating the variables and integrating both sides.

$$\frac{dV}{V} = -\frac{1}{RC}dt$$

Integrating both sides, we get:

$$\ln(V) = -\frac{1}{RC}t + C_1$$

where C1 is a constant of integration.

Applying Initial Conditions

We can apply the initial conditions to find the value of C1. For C1, we have:

$$\ln(5) = -\frac{1}{1 \times 1}t + C_1$$

Solving for C1, we get:

$$C_1 = \ln(5) + t$$

Voltage Across Resistor

The voltage across the resistor is given by:

$$V_R = V_C \times \frac{R}{R+R}$$

where V_C is the voltage across the capacitor.

Substituting the expression for V_C, we get:

$$V_R = (e^{-t} \times 5) \times \frac{1}{2}$$

Simplifying, we get:

$$V_R = \frac{5}{2}e^{-t}$$

Conclusion

In this article, we derived the analytic solution for the voltage across the resistor in a capacitor circuit. We used the differential equation for a capacitor and resistor in series and applied the initial conditions to find the value of the constant of integration. The final expression for the voltage across the resistor is given by:

$$V_R = \frac{5}{2}e^{-t}$$

This solution provides clear understanding of how the voltage across the resistor changes over time in a capacitor circuit.

Mathematical Derivation

Step 1: Separate Variables

The differential equation for a capacitor and resistor in series is given by:

$$\frac{dV}{dt} = -\frac{1}{RC}V$$

We can separate the variables by dividing both sides by V:

$$\frac{dV}{V} = -\frac{1}{RC}dt$$

Step 2: Integrate Both Sides

Integrating both sides, we get:

$$\ln(V) = -\frac{1}{RC}t + C_1$$

where C1 is a constant of integration.

Step 3: Apply Initial Conditions

We can apply the initial conditions to find the value of C1. For C1, we have:

$$\ln(5) = -\frac{1}{1 \times 1}t + C_1$$

Solving for C1, we get:

$$C_1 = \ln(5) + t$$

Step 4: Find Voltage Across Resistor

The voltage across the resistor is given by:

$$V_R = V_C \times \frac{R}{R+R}$$

where V_C is the voltage across the capacitor.

Substituting the expression for V_C, we get:

$$V_R = (e^{-t} \times 5) \times \frac{1}{2}$$

Simplifying, we get:

$$V_R = \frac{5}{2}e^{-t}$$

Code Implementation

Here is a Python code implementation of the analytic solution:

import numpy as np
import matplotlib.pyplot as plt
def voltage_resistor(t):
return 5/2 * np.exp(-t)
t = np.linspace(0, 10, 1000)
v_r = voltage_resistor(t)
plt.plot(t, v_r)
plt.xlabel('Time (s)')
plt.ylabel('Voltage Across Resistor (V)')
plt.title('Voltage Across Resistor Over Time')
plt.show()

This code plots the voltage across the resistor over time, demonstrating the exponential decay of the voltage.

Introduction

In our previous article, we derived the analytic solution for the voltage across the resistor in a capacitor circuit. In this article, we will answer some frequently asked questions related to the topic.

Q: What is the significance of the differential equation for a capacitor and resistor in series?

A: The differential equation for a capacitor and resistor in series is a fundamental equation that describes the behavior of the capacitor and resistor in a circuit. It helps us understand how the voltage across the capacitor changes over time.

Q: How do you separate the variables in the differential equation?

A: To separate the variables, we divide both sides of the differential equation by V, which gives us:

$$\frac{dV}{V} = -\frac{1}{RC}dt$$

Q: What is the role of the constant of integration (C1) in the solution?

A: The constant of integration (C1) is a value that is added to the solution to make it complete. In this case, we used the initial condition to find the value of C1.

Q: How do you apply the initial conditions to find the value of C1?

A: We apply the initial conditions by substituting the initial voltage of the capacitor into the solution. For example, if the initial voltage is 5V, we substitute 5V into the solution to find the value of C1.

Q: What is the final expression for the voltage across the resistor?

A: The final expression for the voltage across the resistor is given by:

$$V_R = \frac{5}{2}e^{-t}$$

Q: How does the voltage across the resistor change over time?

A: The voltage across the resistor decreases exponentially over time, as shown by the expression:

$$V_R = \frac{5}{2}e^{-t}$$

Q: What is the significance of the time constant (RC) in the solution?

A: The time constant (RC) is a measure of how quickly the capacitor charges or discharges. In this case, the time constant is 1 second, which means that the capacitor charges or discharges at a rate of 1 second per second.

Q: How do you plot the voltage across the resistor over time?

A: We can plot the voltage across the resistor over time using a Python code implementation, such as the one shown below:

import numpy as np
import matplotlib.pyplot as plt
def voltage_resistor(t):
return 5/2 * np.exp(-t)
t = np.linspace(0, 10, 1000)
v_r = voltage_resistor(t)
plt.plot(t, v_r)
plt.xlabel('Time (s)')
plt.ylabel('Voltage Across Resistor (V)')
plt.title('Voltage Across Resistor Over Time')
plt.show()

This code plots the voltage across the resistor over time, demonstrating the exponential decay of the voltage.

Conclusion

In this article, we answered some frequently asked questions related to the analytic solution for the voltage across the resistor in a capacitor circuit. We hope that this article has provided a clear understanding of the topic and has helped to clarify any doubts that readers may have had.

Frequently Asked Questions

Q: What is the analytic solution for the voltage across the resistor?

A: The analytic solution for the voltage across the resistor is given by:

$$V_R = \frac{5}{2}e^{-t}$$

Q: How do you separate the variables in the differential equation?

A: To separate the variables, we divide both sides of the differential equation by V, which gives us:

$$\frac{dV}{V} = -\frac{1}{RC}dt$$

Q: What is the role of the constant of integration (C1) in the solution?

A: The constant of integration (C1) is a value that is added to the solution to make it complete. In this case, we used the initial condition to find the value of C1.

Q: How do you apply the initial conditions to find the value of C1?

A: We apply the initial conditions by substituting the initial voltage of the capacitor into the solution. For example, if the initial voltage is 5V, we substitute 5V into the solution to find the value of C1.

Q: What is the final expression for the voltage across the resistor?

A: The final expression for the voltage across the resistor is given by:

$$V_R = \frac{5}{2}e^{-t}$$

Q: How does the voltage across the resistor change over time?

A: The voltage across the resistor decreases exponentially over time, as shown by the expression:

$$V_R = \frac{5}{2}e^{-t}$$

Q: What is the significance of the time constant (RC) in the solution?

A: The time constant (RC) is a measure of how quickly the capacitor charges or discharges. In this case, the time constant is 1 second, which means that the capacitor charges or discharges at a rate of 1 second per second.

Q: How do you plot the voltage across the resistor over time?

A: We can plot the voltage across the resistor over time using a Python code implementation, such as the one shown below:

import numpy as np
import matplotlib.pyplot as plt
def voltage_resistor(t):
return 5/2 * np.exp(-t)
t = np.linspace(0, 10, 1000)
v_r = voltage_resistor(t)
plt.plot(t, v_r)
plt.xlabel('Time (s)')
plt.ylabel('Voltage Across Resistor (V)')
plt.title('Voltage Across Resistor Over Time')
plt.show()

This code plots the voltage across the resistor over time, demonstrating the exponential decay of the voltage.