What is step response of RL series circuit?
2 Answers
The step response of an RL series circuit is the response of the circuit when a step voltage (a sudden change from zero to a constant voltage) is applied to it. Here’s how to analyze it:
1. **Circuit Components**: An RL series circuit consists of a resistor (R) and an inductor (L) connected in series with a voltage source (V).
2. **Differential Equation**: When the step voltage is applied, the voltage across the inductor (L) and resistor (R) can be described by the equation:
\[
V = L \frac{di(t)}{dt} + Ri(t)
\]
where \(i(t)\) is the current through the circuit.
3. **Initial Conditions**: At \(t=0\), when the step is applied, the current \(i(0) = 0\) since the inductor opposes any sudden change in current.
4. **Solving the Equation**: The equation can be rearranged and solved to find the current as a function of time:
\[
\frac{di(t)}{dt} + \frac{R}{L}i(t) = \frac{V}{L}
\]
5. **Time Constant**: The time constant \(\tau\) for the RL circuit is given by:
\[
\tau = \frac{L}{R}
\]
6. **Current Response**: The solution to the differential equation yields the current as:
\[
i(t) = \frac{V}{R}(1 - e^{-\frac{R}{L}t}) \quad \text{for } t \geq 0
\]
This shows that the current rises exponentially from 0 to its steady-state value \( \frac{V}{R} \).
7. **Inductor Voltage**: The voltage across the inductor will initially be equal to the applied voltage and will decay to zero as the current reaches its steady-state value:
\[
V_L(t) = V e^{-\frac{R}{L}t}
\]
8. **Steady-State Values**: Eventually, as \(t \to \infty\):
- Current \(i(t) \to \frac{V}{R}\)
- Voltage across the inductor \(V_L(t) \to 0\)
This response illustrates how the RL circuit reacts over time to a sudden application of voltage, showcasing both the transient and steady-state behavior.
1. **Circuit Components**: An RL series circuit consists of a resistor (R) and an inductor (L) connected in series with a voltage source (V).
2. **Differential Equation**: When the step voltage is applied, the voltage across the inductor (L) and resistor (R) can be described by the equation:
\[
V = L \frac{di(t)}{dt} + Ri(t)
\]
where \(i(t)\) is the current through the circuit.
3. **Initial Conditions**: At \(t=0\), when the step is applied, the current \(i(0) = 0\) since the inductor opposes any sudden change in current.
4. **Solving the Equation**: The equation can be rearranged and solved to find the current as a function of time:
\[
\frac{di(t)}{dt} + \frac{R}{L}i(t) = \frac{V}{L}
\]
5. **Time Constant**: The time constant \(\tau\) for the RL circuit is given by:
\[
\tau = \frac{L}{R}
\]
6. **Current Response**: The solution to the differential equation yields the current as:
\[
i(t) = \frac{V}{R}(1 - e^{-\frac{R}{L}t}) \quad \text{for } t \geq 0
\]
This shows that the current rises exponentially from 0 to its steady-state value \( \frac{V}{R} \).
7. **Inductor Voltage**: The voltage across the inductor will initially be equal to the applied voltage and will decay to zero as the current reaches its steady-state value:
\[
V_L(t) = V e^{-\frac{R}{L}t}
\]
8. **Steady-State Values**: Eventually, as \(t \to \infty\):
- Current \(i(t) \to \frac{V}{R}\)
- Voltage across the inductor \(V_L(t) \to 0\)
This response illustrates how the RL circuit reacts over time to a sudden application of voltage, showcasing both the transient and steady-state behavior.
The step response of an RL series circuit is how the circuit reacts over time when a sudden change, such as a step input (like switching on a voltage source), is applied. Let's break this down to understand it in detail.
### 1. **RL Series Circuit Basics**
An RL series circuit consists of a resistor (R) and an inductor (L) connected in series. When a voltage is applied across the series combination, the behavior of the circuit depends on the resistor and inductor values.
### 2. **Step Input**
A step input refers to a sudden change in voltage. For instance, if the voltage across the circuit jumps from 0V to a certain value (say \( V_0 \)) at time \( t = 0 \), it’s called a step input. This kind of input is used to study how the circuit responds over time.
### 3. **Initial Conditions**
Before \( t = 0 \), let’s assume the circuit is at rest with no current flowing through it and no voltage applied. When the step input is applied at \( t = 0 \), the voltage source suddenly changes from 0V to \( V_0 \).
### 4. **Circuit Behavior**
#### **1. Current Through the Circuit:**
The current \( I(t) \) through an RL series circuit with a step input can be described by the following differential equation:
\[ V(t) = L \frac{dI(t)}{dt} + R I(t) \]
where:
- \( V(t) \) is the applied voltage (which is \( V_0 \) for \( t \geq 0 \)).
- \( L \) is the inductance.
- \( R \) is the resistance.
- \( I(t) \) is the current through the circuit.
For a step input \( V(t) = V_0 \), the differential equation becomes:
\[ V_0 = L \frac{dI(t)}{dt} + R I(t) \]
#### **2. Solution to the Differential Equation:**
To solve this, we can rewrite the equation as:
\[ \frac{dI(t)}{dt} + \frac{R}{L} I(t) = \frac{V_0}{L} \]
This is a first-order linear differential equation. The general solution involves finding the homogeneous solution (related to the circuit's natural response) and a particular solution (related to the forced response due to the step input).
**Homogeneous Solution:**
For \( t < 0 \), the circuit is at equilibrium, so the initial current \( I(0) \) is 0. For \( t \geq 0 \):
\[ I_h(t) = I_{0} e^{-\frac{R}{L} t} \]
where \( I_{0} \) is the initial current (which is zero for \( t = 0 \)).
**Particular Solution:**
Since the input is constant (\( V_0 \)):
\[ I_p(t) = \frac{V_0}{R} \]
**Complete Solution:**
Combining these, the total solution is:
\[ I(t) = \frac{V_0}{R} \left( 1 - e^{-\frac{R}{L} t} \right) \]
Here:
- \( \frac{V_0}{R} \) is the final steady-state current after a long time.
- \( e^{-\frac{R}{L} t} \) represents the transient response, which decays over time.
### 5. **Graphical Representation**
- **At \( t = 0 \):** The current starts from 0A because the inductor initially opposes any change in current.
- **As \( t \) increases:** The current rises exponentially and approaches \( \frac{V_0}{R} \) asymptotically.
- **After a long time:** The current stabilizes at \( \frac{V_0}{R} \), which is the steady-state value.
### 6. **Time Constant**
The rate at which the current approaches its final value is determined by the time constant \( \tau \), given by:
\[ \tau = \frac{L}{R} \]
The time constant indicates how quickly the circuit responds to changes. Specifically, after a time of \( \tau \), the current will have risen to approximately \( 63.2\% \) of its final steady-state value.
### Summary
In an RL series circuit, the step response shows an exponential rise in current from zero to its final steady-state value. This behavior is characterized by a time constant \( \tau \), which dictates how quickly the current reaches its steady-state value after the step input is applied.
### 1. **RL Series Circuit Basics**
An RL series circuit consists of a resistor (R) and an inductor (L) connected in series. When a voltage is applied across the series combination, the behavior of the circuit depends on the resistor and inductor values.
### 2. **Step Input**
A step input refers to a sudden change in voltage. For instance, if the voltage across the circuit jumps from 0V to a certain value (say \( V_0 \)) at time \( t = 0 \), it’s called a step input. This kind of input is used to study how the circuit responds over time.
### 3. **Initial Conditions**
Before \( t = 0 \), let’s assume the circuit is at rest with no current flowing through it and no voltage applied. When the step input is applied at \( t = 0 \), the voltage source suddenly changes from 0V to \( V_0 \).
### 4. **Circuit Behavior**
#### **1. Current Through the Circuit:**
The current \( I(t) \) through an RL series circuit with a step input can be described by the following differential equation:
\[ V(t) = L \frac{dI(t)}{dt} + R I(t) \]
where:
- \( V(t) \) is the applied voltage (which is \( V_0 \) for \( t \geq 0 \)).
- \( L \) is the inductance.
- \( R \) is the resistance.
- \( I(t) \) is the current through the circuit.
For a step input \( V(t) = V_0 \), the differential equation becomes:
\[ V_0 = L \frac{dI(t)}{dt} + R I(t) \]
#### **2. Solution to the Differential Equation:**
To solve this, we can rewrite the equation as:
\[ \frac{dI(t)}{dt} + \frac{R}{L} I(t) = \frac{V_0}{L} \]
This is a first-order linear differential equation. The general solution involves finding the homogeneous solution (related to the circuit's natural response) and a particular solution (related to the forced response due to the step input).
**Homogeneous Solution:**
For \( t < 0 \), the circuit is at equilibrium, so the initial current \( I(0) \) is 0. For \( t \geq 0 \):
\[ I_h(t) = I_{0} e^{-\frac{R}{L} t} \]
where \( I_{0} \) is the initial current (which is zero for \( t = 0 \)).
**Particular Solution:**
Since the input is constant (\( V_0 \)):
\[ I_p(t) = \frac{V_0}{R} \]
**Complete Solution:**
Combining these, the total solution is:
\[ I(t) = \frac{V_0}{R} \left( 1 - e^{-\frac{R}{L} t} \right) \]
Here:
- \( \frac{V_0}{R} \) is the final steady-state current after a long time.
- \( e^{-\frac{R}{L} t} \) represents the transient response, which decays over time.
### 5. **Graphical Representation**
- **At \( t = 0 \):** The current starts from 0A because the inductor initially opposes any change in current.
- **As \( t \) increases:** The current rises exponentially and approaches \( \frac{V_0}{R} \) asymptotically.
- **After a long time:** The current stabilizes at \( \frac{V_0}{R} \), which is the steady-state value.
### 6. **Time Constant**
The rate at which the current approaches its final value is determined by the time constant \( \tau \), given by:
\[ \tau = \frac{L}{R} \]
The time constant indicates how quickly the circuit responds to changes. Specifically, after a time of \( \tau \), the current will have risen to approximately \( 63.2\% \) of its final steady-state value.
### Summary
In an RL series circuit, the step response shows an exponential rise in current from zero to its final steady-state value. This behavior is characterized by a time constant \( \tau \), which dictates how quickly the current reaches its steady-state value after the step input is applied.