What is the response of the RL series circuit?
2 Answers
The response of an RL series circuit can be analyzed in both time domain and frequency domain.
### 1. **Time Domain Response**
In an RL series circuit, the response to a step input voltage \( V(t) \) can be described by the following:
#### **Step Response**
For a step input \( V(t) = V_0 u(t) \) (where \( u(t) \) is the unit step function), the current \( i(t) \) in the RL circuit will be:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
where:
- \( \tau = \frac{L}{R} \) is the time constant of the circuit.
- \( V_0 \) is the magnitude of the step input voltage.
#### **Natural Response**
If the input is removed (i.e., \( V(t) = 0 \)), the current will decay exponentially:
\[ i(t) = I_0 e^{-\frac{t}{\tau}} \]
where \( I_0 \) is the initial current through the inductor.
### 2. **Frequency Domain Response**
In the frequency domain, the impedance \( Z \) of an RL series circuit is given by:
\[ Z = R + j\omega L \]
where \( \omega \) is the angular frequency of the input signal.
#### **Magnitude and Phase Response**
- **Magnitude**: \( |Z| = \sqrt{R^2 + (\omega L)^2} \)
- **Phase**: \( \theta = \arctan\left(\frac{\omega L}{R}\right) \)
The voltage across the resistor \( V_R \) and the inductor \( V_L \) can be found using voltage division:
\[ V_R = V_{\text{in}} \frac{R}{\sqrt{R^2 + (\omega L)^2}} \]
\[ V_L = V_{\text{in}} \frac{\omega L}{\sqrt{R^2 + (\omega L)^2}} \]
where \( V_{\text{in}} \) is the input voltage.
These responses help in understanding how the circuit behaves in different scenarios, such as under a step input or a sinusoidal input at various frequencies.
### 1. **Time Domain Response**
In an RL series circuit, the response to a step input voltage \( V(t) \) can be described by the following:
#### **Step Response**
For a step input \( V(t) = V_0 u(t) \) (where \( u(t) \) is the unit step function), the current \( i(t) \) in the RL circuit will be:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
where:
- \( \tau = \frac{L}{R} \) is the time constant of the circuit.
- \( V_0 \) is the magnitude of the step input voltage.
#### **Natural Response**
If the input is removed (i.e., \( V(t) = 0 \)), the current will decay exponentially:
\[ i(t) = I_0 e^{-\frac{t}{\tau}} \]
where \( I_0 \) is the initial current through the inductor.
### 2. **Frequency Domain Response**
In the frequency domain, the impedance \( Z \) of an RL series circuit is given by:
\[ Z = R + j\omega L \]
where \( \omega \) is the angular frequency of the input signal.
#### **Magnitude and Phase Response**
- **Magnitude**: \( |Z| = \sqrt{R^2 + (\omega L)^2} \)
- **Phase**: \( \theta = \arctan\left(\frac{\omega L}{R}\right) \)
The voltage across the resistor \( V_R \) and the inductor \( V_L \) can be found using voltage division:
\[ V_R = V_{\text{in}} \frac{R}{\sqrt{R^2 + (\omega L)^2}} \]
\[ V_L = V_{\text{in}} \frac{\omega L}{\sqrt{R^2 + (\omega L)^2}} \]
where \( V_{\text{in}} \) is the input voltage.
These responses help in understanding how the circuit behaves in different scenarios, such as under a step input or a sinusoidal input at various frequencies.
In an RL series circuit, which consists of a resistor \( R \) and an inductor \( L \) connected in series, the response refers to how the circuit reacts to various types of input signals. The primary responses of an RL circuit can be categorized based on the type of input signal: step input, sinusoidal input, and transient response.
### 1. **Step Input Response**
When a step input voltage (a sudden change from zero to a constant value) is applied to the RL circuit, the circuit's response can be analyzed in terms of both the voltage across the resistor and the current through the circuit.
- **Current Through the Circuit:**
The current \( I(t) \) in the circuit can be described by the following equation:
\[
I(t) = \frac{V_{\text{step}}}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
where \( V_{\text{step}} \) is the step input voltage, and \( t \) is time. This equation shows that the current starts from zero and asymptotically approaches \( \frac{V_{\text{step}}}{R} \) as \( t \) increases. The time constant of the circuit is \( \tau = \frac{L}{R} \), which characterizes how quickly the current reaches its steady-state value.
- **Voltage Across the Inductor:**
The voltage across the inductor \( V_L(t) \) initially will be equal to the input voltage \( V_{\text{step}} \), but it decays to zero as time progresses:
\[
V_L(t) = V_{\text{step}} e^{-\frac{R}{L}t}
\]
Initially, \( V_L(0) = V_{\text{step}} \), and it exponentially decreases to zero as the current increases.
### 2. **Sinusoidal Input Response**
For a sinusoidal input voltage of the form \( V(t) = V_0 \sin(\omega t) \), where \( \omega \) is the angular frequency of the input signal, the response of the RL circuit can be expressed in terms of its impedance and phase shift.
- **Impedance:**
The impedance \( Z \) of the RL circuit is given by:
\[
Z = \sqrt{R^2 + (\omega L)^2}
\]
where \( \omega = 2\pi f \) is the angular frequency and \( f \) is the frequency of the input signal.
- **Current Through the Circuit:**
The current \( I(t) \) can be found by dividing the input voltage by the impedance:
\[
I(t) = \frac{V_0}{Z} \sin(\omega t - \phi)
\]
where \( \phi \) is the phase angle given by:
\[
\phi = \arctan\left(\frac{\omega L}{R}\right)
\]
- **Voltage Across Each Component:**
- **Voltage Across the Resistor:**
\[
V_R(t) = I(t) \cdot R = \frac{V_0 R}{Z} \sin(\omega t - \phi)
\]
- **Voltage Across the Inductor:**
\[
V_L(t) = I(t) \cdot (\omega L) = \frac{V_0 \omega L}{Z} \sin(\omega t - \phi + \frac{\pi}{2})
\]
### 3. **Transient Response**
The transient response of the RL circuit describes how the circuit behaves when a sudden change (like switching on or off) occurs. This response involves a combination of exponential growth or decay, depending on whether the circuit is being energized or de-energized.
- **When Energizing (Switching On):**
The current grows exponentially from zero according to the step response formula mentioned above.
- **When De-Energizing (Switching Off):**
The current decays exponentially to zero, and the voltage across the inductor changes correspondingly.
In summary, the RL series circuit's response to different inputs involves analyzing the current, voltage across each component, and how these quantities change over time. For step inputs, the responses are exponential, while for sinusoidal inputs, the responses involve impedance and phase shifts.
### 1. **Step Input Response**
When a step input voltage (a sudden change from zero to a constant value) is applied to the RL circuit, the circuit's response can be analyzed in terms of both the voltage across the resistor and the current through the circuit.
- **Current Through the Circuit:**
The current \( I(t) \) in the circuit can be described by the following equation:
\[
I(t) = \frac{V_{\text{step}}}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
where \( V_{\text{step}} \) is the step input voltage, and \( t \) is time. This equation shows that the current starts from zero and asymptotically approaches \( \frac{V_{\text{step}}}{R} \) as \( t \) increases. The time constant of the circuit is \( \tau = \frac{L}{R} \), which characterizes how quickly the current reaches its steady-state value.
- **Voltage Across the Inductor:**
The voltage across the inductor \( V_L(t) \) initially will be equal to the input voltage \( V_{\text{step}} \), but it decays to zero as time progresses:
\[
V_L(t) = V_{\text{step}} e^{-\frac{R}{L}t}
\]
Initially, \( V_L(0) = V_{\text{step}} \), and it exponentially decreases to zero as the current increases.
### 2. **Sinusoidal Input Response**
For a sinusoidal input voltage of the form \( V(t) = V_0 \sin(\omega t) \), where \( \omega \) is the angular frequency of the input signal, the response of the RL circuit can be expressed in terms of its impedance and phase shift.
- **Impedance:**
The impedance \( Z \) of the RL circuit is given by:
\[
Z = \sqrt{R^2 + (\omega L)^2}
\]
where \( \omega = 2\pi f \) is the angular frequency and \( f \) is the frequency of the input signal.
- **Current Through the Circuit:**
The current \( I(t) \) can be found by dividing the input voltage by the impedance:
\[
I(t) = \frac{V_0}{Z} \sin(\omega t - \phi)
\]
where \( \phi \) is the phase angle given by:
\[
\phi = \arctan\left(\frac{\omega L}{R}\right)
\]
- **Voltage Across Each Component:**
- **Voltage Across the Resistor:**
\[
V_R(t) = I(t) \cdot R = \frac{V_0 R}{Z} \sin(\omega t - \phi)
\]
- **Voltage Across the Inductor:**
\[
V_L(t) = I(t) \cdot (\omega L) = \frac{V_0 \omega L}{Z} \sin(\omega t - \phi + \frac{\pi}{2})
\]
### 3. **Transient Response**
The transient response of the RL circuit describes how the circuit behaves when a sudden change (like switching on or off) occurs. This response involves a combination of exponential growth or decay, depending on whether the circuit is being energized or de-energized.
- **When Energizing (Switching On):**
The current grows exponentially from zero according to the step response formula mentioned above.
- **When De-Energizing (Switching Off):**
The current decays exponentially to zero, and the voltage across the inductor changes correspondingly.
In summary, the RL series circuit's response to different inputs involves analyzing the current, voltage across each component, and how these quantities change over time. For step inputs, the responses are exponential, while for sinusoidal inputs, the responses involve impedance and phase shifts.