What is the equation for a first order RL circuit?
2 Answers
A first-order RL circuit is an electrical circuit consisting of a resistor (R) and an inductor (L) connected in series. The behavior of such a circuit can be described by differential equations that relate the voltage, current, and time.
### **Circuit Description**
- **Resistor (R):** A component that opposes the flow of electric current, creating a voltage drop that is proportional to the current.
- **Inductor (L):** A component that stores energy in a magnetic field when current flows through it, and it opposes changes in current by generating a voltage.
### **Key Equations**
#### 1. **Differential Equation for Current (I)**
When a voltage \( V(t) \) is applied to the series RL circuit, the voltage across the resistor \( V_R \) and the inductor \( V_L \) adds up to the total applied voltage:
\[ V(t) = V_R + V_L \]
Using Ohm's Law for the resistor, \( V_R = I(t) \cdot R \), and the voltage across the inductor can be expressed using Faraday's Law of Induction:
\[ V_L = L \frac{dI(t)}{dt} \]
Substitute these into the total voltage equation:
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
Rearranging this equation gives the first-order linear differential equation:
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \]
#### 2. **Solving the Differential Equation**
To solve this equation, we typically look at two scenarios: when the voltage \( V(t) \) is a constant (step input) or a time-varying function.
##### **Step Input Voltage**
Assume the input voltage \( V(t) \) is a step function, i.e., \( V(t) = V_0 \) for \( t \geq 0 \), and the circuit starts with zero current at \( t = 0 \):
The differential equation simplifies to:
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V_0 \]
The solution for the current \( I(t) \) can be found using standard techniques for solving first-order linear differential equations. The general solution is:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
where \( \frac{R}{L} \) is the time constant \( \tau \) of the circuit:
\[ \tau = \frac{L}{R} \]
##### **General Time-Varying Input**
For a general time-varying input \( V(t) \), the solution would involve solving the non-homogeneous differential equation with appropriate methods such as the Laplace transform or variation of parameters, depending on the complexity of \( V(t) \).
### **Summary**
The primary equation for a first-order RL circuit, when subjected to a voltage \( V(t) \), is:
\[ L \frac{dI(t)}{dt} + R I(t) = V(t) \]
This equation describes how the current \( I(t) \) changes over time in response to the applied voltage. The time constant \( \tau = \frac{L}{R} \) is crucial for understanding the transient behavior of the circuit.
### **Circuit Description**
- **Resistor (R):** A component that opposes the flow of electric current, creating a voltage drop that is proportional to the current.
- **Inductor (L):** A component that stores energy in a magnetic field when current flows through it, and it opposes changes in current by generating a voltage.
### **Key Equations**
#### 1. **Differential Equation for Current (I)**
When a voltage \( V(t) \) is applied to the series RL circuit, the voltage across the resistor \( V_R \) and the inductor \( V_L \) adds up to the total applied voltage:
\[ V(t) = V_R + V_L \]
Using Ohm's Law for the resistor, \( V_R = I(t) \cdot R \), and the voltage across the inductor can be expressed using Faraday's Law of Induction:
\[ V_L = L \frac{dI(t)}{dt} \]
Substitute these into the total voltage equation:
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
Rearranging this equation gives the first-order linear differential equation:
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \]
#### 2. **Solving the Differential Equation**
To solve this equation, we typically look at two scenarios: when the voltage \( V(t) \) is a constant (step input) or a time-varying function.
##### **Step Input Voltage**
Assume the input voltage \( V(t) \) is a step function, i.e., \( V(t) = V_0 \) for \( t \geq 0 \), and the circuit starts with zero current at \( t = 0 \):
The differential equation simplifies to:
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V_0 \]
The solution for the current \( I(t) \) can be found using standard techniques for solving first-order linear differential equations. The general solution is:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
where \( \frac{R}{L} \) is the time constant \( \tau \) of the circuit:
\[ \tau = \frac{L}{R} \]
##### **General Time-Varying Input**
For a general time-varying input \( V(t) \), the solution would involve solving the non-homogeneous differential equation with appropriate methods such as the Laplace transform or variation of parameters, depending on the complexity of \( V(t) \).
### **Summary**
The primary equation for a first-order RL circuit, when subjected to a voltage \( V(t) \), is:
\[ L \frac{dI(t)}{dt} + R I(t) = V(t) \]
This equation describes how the current \( I(t) \) changes over time in response to the applied voltage. The time constant \( \tau = \frac{L}{R} \) is crucial for understanding the transient behavior of the circuit.
In a first-order RL circuit, which consists of a resistor (R) and an inductor (L) connected in series, the voltage across the inductor and the resistor follows a particular differential equation. This circuit can be analyzed in both time and frequency domains.
### Time Domain Analysis
When an input voltage \( V_{in}(t) \) is applied to the circuit, the differential equation governing the behavior of the circuit is derived from Kirchhoff's Voltage Law (KVL):
\[ V_{in}(t) = V_R(t) + V_L(t) \]
where:
- \( V_R(t) \) is the voltage across the resistor.
- \( V_L(t) \) is the voltage across the inductor.
For a resistor, the voltage is given by Ohm's Law:
\[ V_R(t) = i(t) \cdot R \]
For an inductor, the voltage is given by:
\[ V_L(t) = L \frac{di(t)}{dt} \]
Combining these, we get:
\[ V_{in}(t) = i(t) \cdot R + L \frac{di(t)}{dt} \]
Rearranging terms to form a standard first-order differential equation:
\[ L \frac{di(t)}{dt} + R i(t) = V_{in}(t) \]
### Solution to the Differential Equation
The solution to this differential equation depends on the nature of \( V_{in}(t) \).
1. **For a Step Input (i.e., \( V_{in}(t) \) is a step function \( V_0 \) applied at \( t = 0 \)):**
The general solution for the current \( i(t) \) can be expressed as:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau = \frac{L}{R} \) is the time constant of the circuit.
2. **For a Sinusoidal Input (i.e., \( V_{in}(t) \) is a sinusoidal function such as \( V_0 \sin(\omega t) \)):**
The response will also be sinusoidal, but with a phase shift and an amplitude determined by the circuit's impedance. The time-domain solution involves complex algebraic manipulation and phasor analysis, which leads to:
\[ i(t) = \frac{V_0}{\sqrt{R^2 + (L\omega)^2}} \sin(\omega t - \phi) \]
where \( \phi \) is the phase angle given by \( \phi = \tan^{-1}\left(\frac{L\omega}{R}\right) \), and \( \omega \) is the angular frequency of the input sinusoid.
### Frequency Domain Analysis
In the frequency domain, you can analyze the circuit using impedances. The impedance of the resistor is \( R \), and the impedance of the inductor is \( j\omega L \), where \( j \) is the imaginary unit and \( \omega \) is the angular frequency.
The total impedance \( Z \) of the RL circuit is:
\[ Z = R + j\omega L \]
The current \( I(\omega) \) in the circuit for a given sinusoidal input \( V_{in}(\omega) \) is:
\[ I(\omega) = \frac{V_{in}(\omega)}{R + j\omega L} \]
This can be used to analyze the steady-state response in the frequency domain.
These analyses provide a comprehensive understanding of the behavior of a first-order RL circuit in both time and frequency domains.
### Time Domain Analysis
When an input voltage \( V_{in}(t) \) is applied to the circuit, the differential equation governing the behavior of the circuit is derived from Kirchhoff's Voltage Law (KVL):
\[ V_{in}(t) = V_R(t) + V_L(t) \]
where:
- \( V_R(t) \) is the voltage across the resistor.
- \( V_L(t) \) is the voltage across the inductor.
For a resistor, the voltage is given by Ohm's Law:
\[ V_R(t) = i(t) \cdot R \]
For an inductor, the voltage is given by:
\[ V_L(t) = L \frac{di(t)}{dt} \]
Combining these, we get:
\[ V_{in}(t) = i(t) \cdot R + L \frac{di(t)}{dt} \]
Rearranging terms to form a standard first-order differential equation:
\[ L \frac{di(t)}{dt} + R i(t) = V_{in}(t) \]
### Solution to the Differential Equation
The solution to this differential equation depends on the nature of \( V_{in}(t) \).
1. **For a Step Input (i.e., \( V_{in}(t) \) is a step function \( V_0 \) applied at \( t = 0 \)):**
The general solution for the current \( i(t) \) can be expressed as:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau = \frac{L}{R} \) is the time constant of the circuit.
2. **For a Sinusoidal Input (i.e., \( V_{in}(t) \) is a sinusoidal function such as \( V_0 \sin(\omega t) \)):**
The response will also be sinusoidal, but with a phase shift and an amplitude determined by the circuit's impedance. The time-domain solution involves complex algebraic manipulation and phasor analysis, which leads to:
\[ i(t) = \frac{V_0}{\sqrt{R^2 + (L\omega)^2}} \sin(\omega t - \phi) \]
where \( \phi \) is the phase angle given by \( \phi = \tan^{-1}\left(\frac{L\omega}{R}\right) \), and \( \omega \) is the angular frequency of the input sinusoid.
### Frequency Domain Analysis
In the frequency domain, you can analyze the circuit using impedances. The impedance of the resistor is \( R \), and the impedance of the inductor is \( j\omega L \), where \( j \) is the imaginary unit and \( \omega \) is the angular frequency.
The total impedance \( Z \) of the RL circuit is:
\[ Z = R + j\omega L \]
The current \( I(\omega) \) in the circuit for a given sinusoidal input \( V_{in}(\omega) \) is:
\[ I(\omega) = \frac{V_{in}(\omega)}{R + j\omega L} \]
This can be used to analyze the steady-state response in the frequency domain.
These analyses provide a comprehensive understanding of the behavior of a first-order RL circuit in both time and frequency domains.