What is the step response of an RL series circuit?
2 Answers
The step response of an RL (Resistor-Inductor) series circuit describes how the circuit reacts to a sudden change in voltage, typically a step input voltage. This response is crucial in understanding the behavior of RL circuits in various applications, such as in filtering and time constant analysis.
### Circuit Description
An RL series circuit consists of a resistor \( R \) and an inductor \( L \) connected in series. When a step input voltage \( V_{in}(t) \) is applied to this circuit, the circuit's response can be analyzed in terms of how the current \( i(t) \) through the circuit changes over time.
### Step Input Voltage
Let's consider the input voltage \( V_{in}(t) \) as a step function, which can be mathematically expressed as:
\[ V_{in}(t) = \begin{cases}
0 & \text{for } t < 0 \\
V_0 & \text{for } t \geq 0
\end{cases} \]
where \( V_0 \) is the magnitude of the step.
### Differential Equation
When the step input voltage is applied, the differential equation governing the RL circuit can be 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 and \( V_L(t) \) is the voltage across the inductor.
Since \( V_R(t) = i(t)R \) and \( V_L(t) = L \frac{di(t)}{dt} \), the equation becomes:
\[ V_0 = i(t)R + L \frac{di(t)}{dt} \]
Rearranging this, we get the first-order linear differential equation:
\[ L \frac{di(t)}{dt} + i(t)R = V_0 \]
### Solution to the Differential Equation
To solve this differential equation, we use the method of integrating factors or recognize it as a standard first-order linear differential equation. The general solution involves finding the homogeneous solution and the particular solution.
1. **Homogeneous Solution:**
For the homogeneous case where \( V_0 = 0 \):
\[ L \frac{di_h(t)}{dt} + i_h(t)R = 0 \]
The solution to this is:
\[ i_h(t) = C e^{-\frac{R}{L}t} \]
where \( C \) is a constant determined by initial conditions.
2. **Particular Solution:**
For the non-homogeneous case where \( V_0 \) is constant, a particular solution can be found as:
\[ i_p(t) = \frac{V_0}{R} \]
3. **Complete Solution:**
Combining both solutions, the total current \( i(t) \) is:
\[ i(t) = i_p(t) + i_h(t) = \frac{V_0}{R} + C e^{-\frac{R}{L}t} \]
### Initial Condition
To determine \( C \), we use the initial condition. At \( t = 0 \), the current \( i(0) \) must be 0 (since the circuit is at rest before the step is applied). Therefore:
\[ 0 = \frac{V_0}{R} + C e^{0} \]
\[ C = -\frac{V_0}{R} \]
So the solution becomes:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
### Summary of Step Response
- **Current \( i(t) \):**
The step response of the RL circuit is given by:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
This shows that the current starts from 0 and asymptotically approaches \( \frac{V_0}{R} \) as \( t \) goes to infinity.
- **Time Constant \( \tau \):**
The time constant \( \tau \) of the RL circuit is:
\[ \tau = \frac{L}{R} \]
The time constant \( \tau \) represents the time it takes for the current to reach approximately 63.2% of its final value \( \frac{V_0}{R} \).
In summary, the step response of an RL series circuit shows an exponential rise in current from zero to \( \frac{V_0}{R} \), characterized by the time constant \( \frac{L}{R} \).
### Circuit Description
An RL series circuit consists of a resistor \( R \) and an inductor \( L \) connected in series. When a step input voltage \( V_{in}(t) \) is applied to this circuit, the circuit's response can be analyzed in terms of how the current \( i(t) \) through the circuit changes over time.
### Step Input Voltage
Let's consider the input voltage \( V_{in}(t) \) as a step function, which can be mathematically expressed as:
\[ V_{in}(t) = \begin{cases}
0 & \text{for } t < 0 \\
V_0 & \text{for } t \geq 0
\end{cases} \]
where \( V_0 \) is the magnitude of the step.
### Differential Equation
When the step input voltage is applied, the differential equation governing the RL circuit can be 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 and \( V_L(t) \) is the voltage across the inductor.
Since \( V_R(t) = i(t)R \) and \( V_L(t) = L \frac{di(t)}{dt} \), the equation becomes:
\[ V_0 = i(t)R + L \frac{di(t)}{dt} \]
Rearranging this, we get the first-order linear differential equation:
\[ L \frac{di(t)}{dt} + i(t)R = V_0 \]
### Solution to the Differential Equation
To solve this differential equation, we use the method of integrating factors or recognize it as a standard first-order linear differential equation. The general solution involves finding the homogeneous solution and the particular solution.
1. **Homogeneous Solution:**
For the homogeneous case where \( V_0 = 0 \):
\[ L \frac{di_h(t)}{dt} + i_h(t)R = 0 \]
The solution to this is:
\[ i_h(t) = C e^{-\frac{R}{L}t} \]
where \( C \) is a constant determined by initial conditions.
2. **Particular Solution:**
For the non-homogeneous case where \( V_0 \) is constant, a particular solution can be found as:
\[ i_p(t) = \frac{V_0}{R} \]
3. **Complete Solution:**
Combining both solutions, the total current \( i(t) \) is:
\[ i(t) = i_p(t) + i_h(t) = \frac{V_0}{R} + C e^{-\frac{R}{L}t} \]
### Initial Condition
To determine \( C \), we use the initial condition. At \( t = 0 \), the current \( i(0) \) must be 0 (since the circuit is at rest before the step is applied). Therefore:
\[ 0 = \frac{V_0}{R} + C e^{0} \]
\[ C = -\frac{V_0}{R} \]
So the solution becomes:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
### Summary of Step Response
- **Current \( i(t) \):**
The step response of the RL circuit is given by:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
This shows that the current starts from 0 and asymptotically approaches \( \frac{V_0}{R} \) as \( t \) goes to infinity.
- **Time Constant \( \tau \):**
The time constant \( \tau \) of the RL circuit is:
\[ \tau = \frac{L}{R} \]
The time constant \( \tau \) represents the time it takes for the current to reach approximately 63.2% of its final value \( \frac{V_0}{R} \).
In summary, the step response of an RL series circuit shows an exponential rise in current from zero to \( \frac{V_0}{R} \), characterized by the time constant \( \frac{L}{R} \).
The step response of an RL series circuit, which consists of a resistor \( R \) and an inductor \( L \) in series, describes how the circuit responds over time when a step input is applied. Here's a detailed breakdown:
### 1. **Circuit Description**
An RL series circuit is composed of:
- A resistor \( R \)
- An inductor \( L \)
When a step input voltage \( V_{in}(t) \) is applied to the circuit, the behavior of the circuit can be analyzed in terms of the voltage across the resistor \( V_R(t) \) and the voltage across the inductor \( V_L(t) \).
### 2. **Applying a Step Input**
Assume we apply a step input voltage \( V_{in}(t) \) that suddenly switches from 0 to \( V_0 \) at \( t = 0 \). Mathematically, the input can be represented as:
\[ V_{in}(t) = V_0 \cdot u(t) \]
where \( u(t) \) is the unit step function.
### 3. **Differential Equation**
The total voltage in the circuit is equal to the sum of the voltages across the resistor and inductor:
\[ V_{in}(t) = V_R(t) + V_L(t) \]
Using Ohm's Law, the voltage across the resistor is:
\[ V_R(t) = i(t) \cdot R \]
The voltage across the inductor is related to the derivative of the current:
\[ V_L(t) = L \cdot \frac{di(t)}{dt} \]
Substitute these into the voltage equation:
\[ V_0 \cdot u(t) = R \cdot i(t) + L \cdot \frac{di(t)}{dt} \]
### 4. **Solving the Differential Equation**
To find the current \( i(t) \), we need to solve the differential equation:
\[ L \cdot \frac{di(t)}{dt} + R \cdot i(t) = V_0 \]
This is a first-order linear differential equation. The solution involves finding the homogeneous solution and the particular solution.
#### Homogeneous Solution:
\[ L \cdot \frac{di_h(t)}{dt} + R \cdot i_h(t) = 0 \]
This can be solved as:
\[ i_h(t) = A \cdot e^{-\frac{R}{L}t} \]
where \( A \) is a constant determined by initial conditions.
#### Particular Solution:
For the particular solution when \( V_{in}(t) = V_0 \):
\[ i_p(t) = \frac{V_0}{R} \]
The general solution is the sum of the homogeneous and particular solutions:
\[ i(t) = i_h(t) + i_p(t) \]
\[ i(t) = A \cdot e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
### 5. **Determining Constants**
At \( t = 0 \), the initial current \( i(0) \) is 0 because the inductor initially opposes the change in current:
\[ i(0) = A + \frac{V_0}{R} = 0 \]
Thus, \( A = -\frac{V_0}{R} \). Therefore:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t} \right) \]
### 6. **Step Response**
The step response for the current \( i(t) \) is:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t} \right) \]
The voltage across the resistor \( V_R(t) \) is:
\[ V_R(t) = i(t) \cdot R = V_0 \left(1 - e^{-\frac{R}{L}t} \right) \]
The voltage across the inductor \( V_L(t) \) is:
\[ V_L(t) = V_{in}(t) - V_R(t) = V_0 \cdot e^{-\frac{R}{L}t} \]
### Summary
The step response of an RL series circuit shows an exponential rise in current \( i(t) \) from 0 to \( \frac{V_0}{R} \) with a time constant \( \tau = \frac{L}{R} \). The voltage across the resistor rises to \( V_0 \) while the voltage across the inductor decays exponentially from \( V_0 \) to 0.
This behavior reflects the inductor’s initial resistance to change in current and its eventual adjustment as the circuit reaches a steady state.
### 1. **Circuit Description**
An RL series circuit is composed of:
- A resistor \( R \)
- An inductor \( L \)
When a step input voltage \( V_{in}(t) \) is applied to the circuit, the behavior of the circuit can be analyzed in terms of the voltage across the resistor \( V_R(t) \) and the voltage across the inductor \( V_L(t) \).
### 2. **Applying a Step Input**
Assume we apply a step input voltage \( V_{in}(t) \) that suddenly switches from 0 to \( V_0 \) at \( t = 0 \). Mathematically, the input can be represented as:
\[ V_{in}(t) = V_0 \cdot u(t) \]
where \( u(t) \) is the unit step function.
### 3. **Differential Equation**
The total voltage in the circuit is equal to the sum of the voltages across the resistor and inductor:
\[ V_{in}(t) = V_R(t) + V_L(t) \]
Using Ohm's Law, the voltage across the resistor is:
\[ V_R(t) = i(t) \cdot R \]
The voltage across the inductor is related to the derivative of the current:
\[ V_L(t) = L \cdot \frac{di(t)}{dt} \]
Substitute these into the voltage equation:
\[ V_0 \cdot u(t) = R \cdot i(t) + L \cdot \frac{di(t)}{dt} \]
### 4. **Solving the Differential Equation**
To find the current \( i(t) \), we need to solve the differential equation:
\[ L \cdot \frac{di(t)}{dt} + R \cdot i(t) = V_0 \]
This is a first-order linear differential equation. The solution involves finding the homogeneous solution and the particular solution.
#### Homogeneous Solution:
\[ L \cdot \frac{di_h(t)}{dt} + R \cdot i_h(t) = 0 \]
This can be solved as:
\[ i_h(t) = A \cdot e^{-\frac{R}{L}t} \]
where \( A \) is a constant determined by initial conditions.
#### Particular Solution:
For the particular solution when \( V_{in}(t) = V_0 \):
\[ i_p(t) = \frac{V_0}{R} \]
The general solution is the sum of the homogeneous and particular solutions:
\[ i(t) = i_h(t) + i_p(t) \]
\[ i(t) = A \cdot e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
### 5. **Determining Constants**
At \( t = 0 \), the initial current \( i(0) \) is 0 because the inductor initially opposes the change in current:
\[ i(0) = A + \frac{V_0}{R} = 0 \]
Thus, \( A = -\frac{V_0}{R} \). Therefore:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t} \right) \]
### 6. **Step Response**
The step response for the current \( i(t) \) is:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t} \right) \]
The voltage across the resistor \( V_R(t) \) is:
\[ V_R(t) = i(t) \cdot R = V_0 \left(1 - e^{-\frac{R}{L}t} \right) \]
The voltage across the inductor \( V_L(t) \) is:
\[ V_L(t) = V_{in}(t) - V_R(t) = V_0 \cdot e^{-\frac{R}{L}t} \]
### Summary
The step response of an RL series circuit shows an exponential rise in current \( i(t) \) from 0 to \( \frac{V_0}{R} \) with a time constant \( \tau = \frac{L}{R} \). The voltage across the resistor rises to \( V_0 \) while the voltage across the inductor decays exponentially from \( V_0 \) to 0.
This behavior reflects the inductor’s initial resistance to change in current and its eventual adjustment as the circuit reaches a steady state.