What is the expression for the RL series circuit?
2 Answers
In an RL series circuit, where a resistor (R) and an inductor (L) are connected in series with a voltage source \( V(t) \), the expression for the circuit’s behavior involves both the resistor and the inductor. Here’s a detailed explanation:
### 1. **Circuit Overview**
The circuit consists of:
- **Resistor (R)**: Opposes current flow and converts electrical energy into heat.
- **Inductor (L)**: Stores energy in its magnetic field and opposes changes in current flow.
### 2. **Voltage and Current Relationship**
Using Kirchhoff's Voltage Law (KVL), the sum of the voltage drops across the resistor and the inductor must equal the applied voltage \( V(t) \).
#### Voltage Across the Resistor
Ohm’s Law gives us the voltage across the resistor:
\[ V_R = I(t) \cdot R \]
where \( I(t) \) is the current through the circuit.
#### Voltage Across the Inductor
The voltage across the inductor is given by:
\[ V_L = L \frac{dI(t)}{dt} \]
where \( \frac{dI(t)}{dt} \) is the rate of change of current with respect to time.
### 3. **Total Voltage Equation**
According to KVL:
\[ V(t) = V_R + V_L \]
Substitute the expressions for \( V_R \) and \( V_L \):
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
### 4. **Differential Equation**
Rearranging this equation to form a differential equation for the current \( I(t) \):
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \]
This is a first-order linear differential equation.
### 5. **Solution to the Differential Equation**
To solve this differential equation, we need to consider two cases: **homogeneous** and **particular** solutions.
#### a. **Homogeneous Solution**
For the homogeneous case where \( V(t) = 0 \):
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = 0 \]
Rearrange to:
\[ \frac{dI(t)}{dt} = -\frac{R}{L} I(t) \]
This is a separable differential equation. Integrating, we get:
\[ I(t) = I_0 e^{-\frac{R}{L} t} \]
where \( I_0 \) is the initial current at \( t = 0 \).
#### b. **Particular Solution**
If \( V(t) \) is a constant \( V_0 \), the particular solution is found by assuming a steady-state current \( I_{ss} \):
\[ L \frac{dI_{ss}}{dt} = 0 \]
\[ I_{ss} = \frac{V_0}{R} \]
So the particular solution is:
\[ I(t) = \frac{V_0}{R} \]
#### c. **General Solution**
The general solution combines both homogeneous and particular solutions:
\[ I(t) = I_{ss} + (I_0 - I_{ss}) e^{-\frac{R}{L} t} \]
where \( I_{ss} = \frac{V(t)}{R} \) for a constant \( V(t) \).
### 6. **Example with a Step Input**
If the input voltage \( V(t) \) is a step function \( V(t) = V_0 \) (i.e., \( V(t) \) jumps from 0 to \( V_0 \) at \( t = 0 \)), the current \( I(t) \) evolves as:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t} \right) \]
### 7. **Frequency Domain Analysis**
In the frequency domain, using Laplace transforms, the impedance of the RL circuit is:
\[ Z = R + j \omega L \]
where \( \omega \) is the angular frequency of the source. The voltage-current relationship in the frequency domain becomes:
\[ V(s) = I(s) \cdot (R + sL) \]
### Summary
The RL series circuit’s behavior is governed by the differential equation \( L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \). The solution involves finding both homogeneous and particular solutions, depending on the nature of the input voltage \( V(t) \).
### 1. **Circuit Overview**
The circuit consists of:
- **Resistor (R)**: Opposes current flow and converts electrical energy into heat.
- **Inductor (L)**: Stores energy in its magnetic field and opposes changes in current flow.
### 2. **Voltage and Current Relationship**
Using Kirchhoff's Voltage Law (KVL), the sum of the voltage drops across the resistor and the inductor must equal the applied voltage \( V(t) \).
#### Voltage Across the Resistor
Ohm’s Law gives us the voltage across the resistor:
\[ V_R = I(t) \cdot R \]
where \( I(t) \) is the current through the circuit.
#### Voltage Across the Inductor
The voltage across the inductor is given by:
\[ V_L = L \frac{dI(t)}{dt} \]
where \( \frac{dI(t)}{dt} \) is the rate of change of current with respect to time.
### 3. **Total Voltage Equation**
According to KVL:
\[ V(t) = V_R + V_L \]
Substitute the expressions for \( V_R \) and \( V_L \):
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
### 4. **Differential Equation**
Rearranging this equation to form a differential equation for the current \( I(t) \):
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \]
This is a first-order linear differential equation.
### 5. **Solution to the Differential Equation**
To solve this differential equation, we need to consider two cases: **homogeneous** and **particular** solutions.
#### a. **Homogeneous Solution**
For the homogeneous case where \( V(t) = 0 \):
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = 0 \]
Rearrange to:
\[ \frac{dI(t)}{dt} = -\frac{R}{L} I(t) \]
This is a separable differential equation. Integrating, we get:
\[ I(t) = I_0 e^{-\frac{R}{L} t} \]
where \( I_0 \) is the initial current at \( t = 0 \).
#### b. **Particular Solution**
If \( V(t) \) is a constant \( V_0 \), the particular solution is found by assuming a steady-state current \( I_{ss} \):
\[ L \frac{dI_{ss}}{dt} = 0 \]
\[ I_{ss} = \frac{V_0}{R} \]
So the particular solution is:
\[ I(t) = \frac{V_0}{R} \]
#### c. **General Solution**
The general solution combines both homogeneous and particular solutions:
\[ I(t) = I_{ss} + (I_0 - I_{ss}) e^{-\frac{R}{L} t} \]
where \( I_{ss} = \frac{V(t)}{R} \) for a constant \( V(t) \).
### 6. **Example with a Step Input**
If the input voltage \( V(t) \) is a step function \( V(t) = V_0 \) (i.e., \( V(t) \) jumps from 0 to \( V_0 \) at \( t = 0 \)), the current \( I(t) \) evolves as:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t} \right) \]
### 7. **Frequency Domain Analysis**
In the frequency domain, using Laplace transforms, the impedance of the RL circuit is:
\[ Z = R + j \omega L \]
where \( \omega \) is the angular frequency of the source. The voltage-current relationship in the frequency domain becomes:
\[ V(s) = I(s) \cdot (R + sL) \]
### Summary
The RL series circuit’s behavior is governed by the differential equation \( L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \). The solution involves finding both homogeneous and particular solutions, depending on the nature of the input voltage \( V(t) \).
In an RL series circuit, the resistor (R) and inductor (L) are connected in series with a voltage source. The key parameters of the circuit are the resistance \( R \), the inductance \( L \), and the voltage source \( V(t) \).
### **1. ** **General Expression:**
The differential equation governing the RL circuit can be derived from Kirchhoff’s Voltage Law (KVL). According to KVL, the sum of the voltage drops across the resistor and the inductor equals the applied voltage. Mathematically:
\[ V(t) = V_R(t) + V_L(t) \]
where:
- \( V(t) \) is the applied voltage across the series combination of R and L.
- \( V_R(t) = i(t) \cdot R \) is the voltage drop across the resistor.
- \( V_L(t) = L \frac{di(t)}{dt} \) is the voltage drop across the inductor.
So the equation becomes:
\[ V(t) = i(t) \cdot R + L \frac{di(t)}{dt} \]
Rearranging this equation gives:
\[ L \frac{di(t)}{dt} + i(t) \cdot R = V(t) \]
### **2. ** **For Different Types of Sources:**
#### **a. ** **DC Voltage Source:**
For a constant DC voltage \( V_0 \) applied at \( t = 0 \), the differential equation is:
\[ L \frac{di(t)}{dt} + R i(t) = V_0 \]
**Solving for \( i(t) \):**
1. **Initial Condition:** At \( t = 0 \), the current is \( i(0) \). Typically, if the circuit has been at rest before \( t = 0 \), \( i(0) = 0 \).
2. **Homogeneous Solution:**
Solve the homogeneous equation \( L \frac{di(t)}{dt} + R i(t) = 0 \):
\[ i_h(t) = A e^{-\frac{R}{L}t} \]
3. **Particular Solution:**
For a constant voltage \( V_0 \), the particular solution is:
\[ i_p(t) = \frac{V_0}{R} \]
4. **General Solution:**
Combine the homogeneous and particular solutions:
\[ i(t) = \frac{V_0}{R} + \left(i(0) - \frac{V_0}{R}\right) e^{-\frac{R}{L}t} \]
If \( i(0) = 0 \), then:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
#### **b. ** **AC Voltage Source:**
For an AC source \( V(t) = V_0 \sin(\omega t) \), where \( \omega \) is the angular frequency:
1. **The Differential Equation:**
\[ L \frac{di(t)}{dt} + R i(t) = V_0 \sin(\omega t) \]
2. **Solving:**
For AC analysis, it's often useful to use phasor analysis. Represent the voltage and current as phasors:
\[ V(t) = V_0 \sin(\omega t) \]
Convert to the phasor domain:
\[ \tilde{V} = V_0 \angle 0^\circ \]
\[ \tilde{I} = \frac{\tilde{V}}{R + j \omega L} \]
The impedance \( Z \) of the RL circuit is:
\[ Z = R + j \omega L \]
So the phasor current is:
\[ \tilde{I} = \frac{V_0}{R + j \omega L} \]
Converting back to the time domain:
\[ i(t) = \frac{V_0}{\sqrt{R^2 + (\omega L)^2}} \sin\left(\omega t - \phi\right) \]
where \( \phi \) is the phase angle:
\[ \tan(\phi) = \frac{\omega L}{R} \]
### **Summary:**
- For a DC source, the current \( i(t) \) rises from zero to a steady-state value \( \frac{V_0}{R} \) with a time constant \( \tau = \frac{L}{R} \).
- For an AC source, the current is a sinusoidal function with the same frequency as the voltage but with a phase shift \( \phi \) and amplitude depending on the impedance of the RL circuit.
These expressions cover the behavior of an RL circuit for both DC and AC sources.
### **1. ** **General Expression:**
The differential equation governing the RL circuit can be derived from Kirchhoff’s Voltage Law (KVL). According to KVL, the sum of the voltage drops across the resistor and the inductor equals the applied voltage. Mathematically:
\[ V(t) = V_R(t) + V_L(t) \]
where:
- \( V(t) \) is the applied voltage across the series combination of R and L.
- \( V_R(t) = i(t) \cdot R \) is the voltage drop across the resistor.
- \( V_L(t) = L \frac{di(t)}{dt} \) is the voltage drop across the inductor.
So the equation becomes:
\[ V(t) = i(t) \cdot R + L \frac{di(t)}{dt} \]
Rearranging this equation gives:
\[ L \frac{di(t)}{dt} + i(t) \cdot R = V(t) \]
### **2. ** **For Different Types of Sources:**
#### **a. ** **DC Voltage Source:**
For a constant DC voltage \( V_0 \) applied at \( t = 0 \), the differential equation is:
\[ L \frac{di(t)}{dt} + R i(t) = V_0 \]
**Solving for \( i(t) \):**
1. **Initial Condition:** At \( t = 0 \), the current is \( i(0) \). Typically, if the circuit has been at rest before \( t = 0 \), \( i(0) = 0 \).
2. **Homogeneous Solution:**
Solve the homogeneous equation \( L \frac{di(t)}{dt} + R i(t) = 0 \):
\[ i_h(t) = A e^{-\frac{R}{L}t} \]
3. **Particular Solution:**
For a constant voltage \( V_0 \), the particular solution is:
\[ i_p(t) = \frac{V_0}{R} \]
4. **General Solution:**
Combine the homogeneous and particular solutions:
\[ i(t) = \frac{V_0}{R} + \left(i(0) - \frac{V_0}{R}\right) e^{-\frac{R}{L}t} \]
If \( i(0) = 0 \), then:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
#### **b. ** **AC Voltage Source:**
For an AC source \( V(t) = V_0 \sin(\omega t) \), where \( \omega \) is the angular frequency:
1. **The Differential Equation:**
\[ L \frac{di(t)}{dt} + R i(t) = V_0 \sin(\omega t) \]
2. **Solving:**
For AC analysis, it's often useful to use phasor analysis. Represent the voltage and current as phasors:
\[ V(t) = V_0 \sin(\omega t) \]
Convert to the phasor domain:
\[ \tilde{V} = V_0 \angle 0^\circ \]
\[ \tilde{I} = \frac{\tilde{V}}{R + j \omega L} \]
The impedance \( Z \) of the RL circuit is:
\[ Z = R + j \omega L \]
So the phasor current is:
\[ \tilde{I} = \frac{V_0}{R + j \omega L} \]
Converting back to the time domain:
\[ i(t) = \frac{V_0}{\sqrt{R^2 + (\omega L)^2}} \sin\left(\omega t - \phi\right) \]
where \( \phi \) is the phase angle:
\[ \tan(\phi) = \frac{\omega L}{R} \]
### **Summary:**
- For a DC source, the current \( i(t) \) rises from zero to a steady-state value \( \frac{V_0}{R} \) with a time constant \( \tau = \frac{L}{R} \).
- For an AC source, the current is a sinusoidal function with the same frequency as the voltage but with a phase shift \( \phi \) and amplitude depending on the impedance of the RL circuit.
These expressions cover the behavior of an RL circuit for both DC and AC sources.