What is the first order differential equation for RL circuits?
2 Answers
In an RL circuit, which consists of a resistor (R) and an inductor (L) connected in series with a voltage source, the behavior of the current flowing through the circuit can be described using a first-order differential equation. Let’s break down the concepts step-by-step to understand how this equation is derived.
### Components of an RL Circuit
1. **Resistor (R)**: A component that opposes the flow of electric current, resulting in a voltage drop across it.
2. **Inductor (L)**: A coil of wire that stores energy in a magnetic field when current passes through it. The inductor opposes changes in current flow.
### Kirchhoff’s Voltage Law
According to **Kirchhoff's Voltage Law**, the sum of the potential differences (voltage) around any closed loop in a circuit must equal zero. For an RL circuit, if we denote:
- \( V \) as the voltage of the source,
- \( i(t) \) as the current flowing through the circuit at time \( t \),
- \( R \) as the resistance in ohms,
- \( L \) as the inductance in henries,
the voltage drop across the resistor and the inductor can be expressed as:
- **Voltage across the resistor**: \( V_R = R \cdot i(t) \)
- **Voltage across the inductor**: \( V_L = L \frac{di(t)}{dt} \)
### Formulating the Differential Equation
Applying Kirchhoff's Voltage Law to the circuit, we have:
\[
V - V_R - V_L = 0
\]
Substituting the expressions for the voltage drops gives:
\[
V - R \cdot i(t) - L \frac{di(t)}{dt} = 0
\]
Rearranging this equation leads to:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V
\]
This is a **first-order linear differential equation** with respect to the current \( i(t) \).
### Standard Form
To express it in standard form, we can write it as:
\[
\frac{di(t)}{dt} + \frac{R}{L} i(t) = \frac{V}{L}
\]
Where:
- \( \frac{R}{L} \) is a constant that represents the rate of change of current with respect to time, and
- \( \frac{V}{L} \) is the driving force of the circuit.
### Solving the Differential Equation
To solve this first-order linear differential equation, we can use methods such as the **integrating factor method**. The general solution involves finding a particular solution and the homogeneous solution.
#### Homogeneous Solution
The homogeneous part of the equation is given by:
\[
\frac{di(t)}{dt} + \frac{R}{L} i(t) = 0
\]
This can be solved using separation of variables:
1. Rearranging gives:
\[
\frac{di(t)}{i(t)} = -\frac{R}{L} dt
\]
2. Integrating both sides leads to:
\[
\ln|i(t)| = -\frac{R}{L} t + C
\]
where \( C \) is the constant of integration.
3. Exponentiating gives:
\[
i(t) = Ce^{-\frac{R}{L} t}
\]
#### Particular Solution
For the particular solution, we can assume \( i(t) \) approaches a steady-state value \( I \) as \( t \) approaches infinity. Setting \( \frac{di(t)}{dt} = 0 \) in the original equation gives:
\[
R \cdot I = V \implies I = \frac{V}{R}
\]
### General Solution
The general solution of the differential equation is a combination of the homogeneous and particular solutions:
\[
i(t) = \frac{V}{R} + Ce^{-\frac{R}{L} t}
\]
Where \( C \) can be determined based on the initial conditions of the circuit (for example, the initial current at \( t = 0 \)).
### Summary
The first-order differential equation governing the current in an RL circuit is:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V
\]
Or in standard form:
\[
\frac{di(t)}{dt} + \frac{R}{L} i(t) = \frac{V}{L}
\]
This equation captures the dynamics of current change in response to an applied voltage in the circuit, illustrating how the inductor opposes changes in current and how resistance affects the flow of current over time.
### Components of an RL Circuit
1. **Resistor (R)**: A component that opposes the flow of electric current, resulting in a voltage drop across it.
2. **Inductor (L)**: A coil of wire that stores energy in a magnetic field when current passes through it. The inductor opposes changes in current flow.
### Kirchhoff’s Voltage Law
According to **Kirchhoff's Voltage Law**, the sum of the potential differences (voltage) around any closed loop in a circuit must equal zero. For an RL circuit, if we denote:
- \( V \) as the voltage of the source,
- \( i(t) \) as the current flowing through the circuit at time \( t \),
- \( R \) as the resistance in ohms,
- \( L \) as the inductance in henries,
the voltage drop across the resistor and the inductor can be expressed as:
- **Voltage across the resistor**: \( V_R = R \cdot i(t) \)
- **Voltage across the inductor**: \( V_L = L \frac{di(t)}{dt} \)
### Formulating the Differential Equation
Applying Kirchhoff's Voltage Law to the circuit, we have:
\[
V - V_R - V_L = 0
\]
Substituting the expressions for the voltage drops gives:
\[
V - R \cdot i(t) - L \frac{di(t)}{dt} = 0
\]
Rearranging this equation leads to:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V
\]
This is a **first-order linear differential equation** with respect to the current \( i(t) \).
### Standard Form
To express it in standard form, we can write it as:
\[
\frac{di(t)}{dt} + \frac{R}{L} i(t) = \frac{V}{L}
\]
Where:
- \( \frac{R}{L} \) is a constant that represents the rate of change of current with respect to time, and
- \( \frac{V}{L} \) is the driving force of the circuit.
### Solving the Differential Equation
To solve this first-order linear differential equation, we can use methods such as the **integrating factor method**. The general solution involves finding a particular solution and the homogeneous solution.
#### Homogeneous Solution
The homogeneous part of the equation is given by:
\[
\frac{di(t)}{dt} + \frac{R}{L} i(t) = 0
\]
This can be solved using separation of variables:
1. Rearranging gives:
\[
\frac{di(t)}{i(t)} = -\frac{R}{L} dt
\]
2. Integrating both sides leads to:
\[
\ln|i(t)| = -\frac{R}{L} t + C
\]
where \( C \) is the constant of integration.
3. Exponentiating gives:
\[
i(t) = Ce^{-\frac{R}{L} t}
\]
#### Particular Solution
For the particular solution, we can assume \( i(t) \) approaches a steady-state value \( I \) as \( t \) approaches infinity. Setting \( \frac{di(t)}{dt} = 0 \) in the original equation gives:
\[
R \cdot I = V \implies I = \frac{V}{R}
\]
### General Solution
The general solution of the differential equation is a combination of the homogeneous and particular solutions:
\[
i(t) = \frac{V}{R} + Ce^{-\frac{R}{L} t}
\]
Where \( C \) can be determined based on the initial conditions of the circuit (for example, the initial current at \( t = 0 \)).
### Summary
The first-order differential equation governing the current in an RL circuit is:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V
\]
Or in standard form:
\[
\frac{di(t)}{dt} + \frac{R}{L} i(t) = \frac{V}{L}
\]
This equation captures the dynamics of current change in response to an applied voltage in the circuit, illustrating how the inductor opposes changes in current and how resistance affects the flow of current over time.
The first-order differential equation for an RL circuit (a circuit with a resistor \(R\) and an inductor \(L\)) describes the relationship between the voltage across the circuit components and the current flowing through them. To derive this equation, consider a simple series RL circuit, where a resistor \(R\) and an inductor \(L\) are connected in series with a voltage source \(V(t)\).
### Derivation of the Differential Equation
1. **Identify Components and Variables:**
- Let \( i(t) \) be the current flowing through the circuit.
- The voltage across the resistor is \( V_R(t) = R \cdot i(t) \).
- The voltage across the inductor is \( V_L(t) = L \frac{di(t)}{dt} \), where \( \frac{di(t)}{dt} \) is the rate of change of current with respect to time.
2. **Apply Kirchhoff’s Voltage Law (KVL):**
Kirchhoff's Voltage Law states that the sum of all voltages around a closed loop is zero. For the series RL circuit, the total voltage provided by the source \( V(t) \) equals the sum of the voltages across the resistor and the inductor:
\[
V(t) = V_R(t) + V_L(t)
\]
Substituting the expressions for \( V_R(t) \) and \( V_L(t) \):
\[
V(t) = R \cdot i(t) + L \frac{di(t)}{dt}
\]
3. **Form the Differential Equation:**
Rearranging the terms to isolate the derivative of the current:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V(t)
\]
This is the first-order linear differential equation that describes the behavior of the current in an RL circuit.
### Explanation of the Differential Equation
- **\(L \frac{di(t)}{dt}\)**: This term represents the voltage drop across the inductor. The inductance \(L\) influences how quickly the current changes over time.
- **\(R \cdot i(t)\)**: This term represents the voltage drop across the resistor. The resistance \(R\) directly affects the amount of voltage drop for a given current.
- **\(V(t)\)**: This is the external voltage applied to the circuit. It drives the current through the resistor and inductor.
### Special Cases
1. **Steady-State Response**: When \( V(t) \) is constant (DC source), the differential equation simplifies. For a constant voltage \( V \), the current \( i(t) \) will eventually reach a steady-state value where \( \frac{di(t)}{dt} = 0 \):
\[
R \cdot i(t) = V \implies i(t) = \frac{V}{R}
\]
2. **Transient Response**: For time-varying \( V(t) \), the differential equation can be solved using techniques for first-order linear differential equations, depending on the specific form of \( V(t) \).
This differential equation is fundamental in understanding how the RL circuit responds to different types of input voltages over time.
### Derivation of the Differential Equation
1. **Identify Components and Variables:**
- Let \( i(t) \) be the current flowing through the circuit.
- The voltage across the resistor is \( V_R(t) = R \cdot i(t) \).
- The voltage across the inductor is \( V_L(t) = L \frac{di(t)}{dt} \), where \( \frac{di(t)}{dt} \) is the rate of change of current with respect to time.
2. **Apply Kirchhoff’s Voltage Law (KVL):**
Kirchhoff's Voltage Law states that the sum of all voltages around a closed loop is zero. For the series RL circuit, the total voltage provided by the source \( V(t) \) equals the sum of the voltages across the resistor and the inductor:
\[
V(t) = V_R(t) + V_L(t)
\]
Substituting the expressions for \( V_R(t) \) and \( V_L(t) \):
\[
V(t) = R \cdot i(t) + L \frac{di(t)}{dt}
\]
3. **Form the Differential Equation:**
Rearranging the terms to isolate the derivative of the current:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V(t)
\]
This is the first-order linear differential equation that describes the behavior of the current in an RL circuit.
### Explanation of the Differential Equation
- **\(L \frac{di(t)}{dt}\)**: This term represents the voltage drop across the inductor. The inductance \(L\) influences how quickly the current changes over time.
- **\(R \cdot i(t)\)**: This term represents the voltage drop across the resistor. The resistance \(R\) directly affects the amount of voltage drop for a given current.
- **\(V(t)\)**: This is the external voltage applied to the circuit. It drives the current through the resistor and inductor.
### Special Cases
1. **Steady-State Response**: When \( V(t) \) is constant (DC source), the differential equation simplifies. For a constant voltage \( V \), the current \( i(t) \) will eventually reach a steady-state value where \( \frac{di(t)}{dt} = 0 \):
\[
R \cdot i(t) = V \implies i(t) = \frac{V}{R}
\]
2. **Transient Response**: For time-varying \( V(t) \), the differential equation can be solved using techniques for first-order linear differential equations, depending on the specific form of \( V(t) \).
This differential equation is fundamental in understanding how the RL circuit responds to different types of input voltages over time.