How do you derive current in an RL circuit?
2 Answers
To derive the current in an RL (resistor-inductor) circuit, you start with Kirchhoff's voltage law. In a simple series RL circuit with a voltage source \( V \), a resistor \( R \), and an inductor \( L \), the equation can be set up as follows:
1. **Apply Kirchhoff’s Voltage Law**:
\[
V = V_R + V_L
\]
where \( V_R = iR \) (voltage across the resistor) and \( V_L = L \frac{di}{dt} \) (voltage across the inductor).
Thus, the equation becomes:
\[
V = iR + L \frac{di}{dt}
\]
2. **Rearrange the equation**:
\[
L \frac{di}{dt} + iR = V
\]
3. **Separate variables**:
Rearranging gives:
\[
L \frac{di}{dt} = V - iR
\]
or
\[
\frac{di}{V - iR} = \frac{dt}{L}
\]
4. **Integrate both sides**:
Integrating the left side with respect to \( i \) and the right side with respect to \( t \):
\[
\int \frac{di}{V - iR} = \int \frac{dt}{L}
\]
The left integral yields:
\[
-\frac{1}{R} \ln |V - iR| = \frac{t}{L} + C
\]
where \( C \) is the constant of integration.
5. **Solve for \( i \)**:
Rearranging gives:
\[
|V - iR| = e^{-Rt/L}(e^{-RC}L)
\]
Assuming the initial condition where \( i(0) = 0 \) (the current starts from zero), you can solve for \( C \) and get the specific form of the current.
6. **Final equation**:
The solution for the current \( i(t) \) in an RL circuit is:
\[
i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
This equation describes how the current builds up over time in an RL circuit, approaching its steady-state value \( \frac{V}{R} \) as \( t \) increases. The time constant \( \tau = \frac{L}{R} \) characterizes the rate of this growth.
1. **Apply Kirchhoff’s Voltage Law**:
\[
V = V_R + V_L
\]
where \( V_R = iR \) (voltage across the resistor) and \( V_L = L \frac{di}{dt} \) (voltage across the inductor).
Thus, the equation becomes:
\[
V = iR + L \frac{di}{dt}
\]
2. **Rearrange the equation**:
\[
L \frac{di}{dt} + iR = V
\]
3. **Separate variables**:
Rearranging gives:
\[
L \frac{di}{dt} = V - iR
\]
or
\[
\frac{di}{V - iR} = \frac{dt}{L}
\]
4. **Integrate both sides**:
Integrating the left side with respect to \( i \) and the right side with respect to \( t \):
\[
\int \frac{di}{V - iR} = \int \frac{dt}{L}
\]
The left integral yields:
\[
-\frac{1}{R} \ln |V - iR| = \frac{t}{L} + C
\]
where \( C \) is the constant of integration.
5. **Solve for \( i \)**:
Rearranging gives:
\[
|V - iR| = e^{-Rt/L}(e^{-RC}L)
\]
Assuming the initial condition where \( i(0) = 0 \) (the current starts from zero), you can solve for \( C \) and get the specific form of the current.
6. **Final equation**:
The solution for the current \( i(t) \) in an RL circuit is:
\[
i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
This equation describes how the current builds up over time in an RL circuit, approaching its steady-state value \( \frac{V}{R} \) as \( t \) increases. The time constant \( \tau = \frac{L}{R} \) characterizes the rate of this growth.
To derive the current in an RL circuit (a resistor-inductor circuit) in both the transient and steady-state conditions, you can follow these steps:
### 1. **RL Circuit Description:**
An RL circuit typically consists of a resistor \( R \) and an inductor \( L \) connected in series with a voltage source \( V(t) \).
### 2. **Differential Equation:**
The voltage across the resistor \( V_R \) and the inductor \( V_L \) can be described using Kirchhoff's Voltage Law (KVL):
\[ V(t) = V_R + V_L \]
The voltage across the resistor is given by Ohm's Law:
\[ V_R = i(t)R \]
The voltage across the inductor is related to the rate of change of current:
\[ V_L = L \frac{di(t)}{dt} \]
So, substituting these into the KVL equation:
\[ V(t) = i(t)R + L \frac{di(t)}{dt} \]
Rearrange this to form a differential equation:
\[ L \frac{di(t)}{dt} + i(t)R = V(t) \]
### 3. **Solving the Differential Equation:**
#### **a. Homogeneous Solution (Zero Input):**
First, solve the homogeneous differential equation where \( V(t) = 0 \):
\[ L \frac{di(t)}{dt} + i(t)R = 0 \]
Rearrange to:
\[ \frac{di(t)}{dt} = -\frac{R}{L} i(t) \]
This is a first-order linear differential equation with the solution:
\[ i_h(t) = A e^{-\frac{R}{L}t} \]
where \( A \) is a constant determined by initial conditions.
#### **b. Particular Solution (Non-zero Input):**
For a constant voltage \( V(t) = V_0 \):
Substitute \( V(t) = V_0 \) into the differential equation:
\[ L \frac{di(t)}{dt} + i(t)R = V_0 \]
Assume a particular solution of the form \( i_p(t) = I_0 \) (a constant):
Substitute \( i_p(t) = I_0 \) into the differential equation:
\[ L \cdot 0 + I_0 R = V_0 \]
Solve for \( I_0 \):
\[ I_0 = \frac{V_0}{R} \]
#### **c. General Solution:**
The general solution is the sum of the homogeneous and particular solutions:
\[ i(t) = i_h(t) + i_p(t) \]
\[ i(t) = A e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
#### **d. Applying Initial Conditions:**
Typically, the initial current \( i(0) \) is used to find the constant \( A \). If the initial current is zero (for a switch closed at \( t = 0 \)):
\[ i(0) = A + \frac{V_0}{R} \]
Solving for \( A \):
\[ A = i(0) - \frac{V_0}{R} \]
So, the current as a function of time is:
\[ i(t) = \left(i(0) - \frac{V_0}{R}\right) e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
### **Summary:**
- **Transient Response:** \( i(t) = \left(i(0) - \frac{V_0}{R}\right) e^{-\frac{R}{L}t} + \frac{V_0}{R} \)
- **Steady-State Response:** \( i(t) \to \frac{V_0}{R} \) as \( t \to \infty \)
This formula describes how the current changes over time after a voltage is applied to the RL circuit.
### 1. **RL Circuit Description:**
An RL circuit typically consists of a resistor \( R \) and an inductor \( L \) connected in series with a voltage source \( V(t) \).
### 2. **Differential Equation:**
The voltage across the resistor \( V_R \) and the inductor \( V_L \) can be described using Kirchhoff's Voltage Law (KVL):
\[ V(t) = V_R + V_L \]
The voltage across the resistor is given by Ohm's Law:
\[ V_R = i(t)R \]
The voltage across the inductor is related to the rate of change of current:
\[ V_L = L \frac{di(t)}{dt} \]
So, substituting these into the KVL equation:
\[ V(t) = i(t)R + L \frac{di(t)}{dt} \]
Rearrange this to form a differential equation:
\[ L \frac{di(t)}{dt} + i(t)R = V(t) \]
### 3. **Solving the Differential Equation:**
#### **a. Homogeneous Solution (Zero Input):**
First, solve the homogeneous differential equation where \( V(t) = 0 \):
\[ L \frac{di(t)}{dt} + i(t)R = 0 \]
Rearrange to:
\[ \frac{di(t)}{dt} = -\frac{R}{L} i(t) \]
This is a first-order linear differential equation with the solution:
\[ i_h(t) = A e^{-\frac{R}{L}t} \]
where \( A \) is a constant determined by initial conditions.
#### **b. Particular Solution (Non-zero Input):**
For a constant voltage \( V(t) = V_0 \):
Substitute \( V(t) = V_0 \) into the differential equation:
\[ L \frac{di(t)}{dt} + i(t)R = V_0 \]
Assume a particular solution of the form \( i_p(t) = I_0 \) (a constant):
Substitute \( i_p(t) = I_0 \) into the differential equation:
\[ L \cdot 0 + I_0 R = V_0 \]
Solve for \( I_0 \):
\[ I_0 = \frac{V_0}{R} \]
#### **c. General Solution:**
The general solution is the sum of the homogeneous and particular solutions:
\[ i(t) = i_h(t) + i_p(t) \]
\[ i(t) = A e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
#### **d. Applying Initial Conditions:**
Typically, the initial current \( i(0) \) is used to find the constant \( A \). If the initial current is zero (for a switch closed at \( t = 0 \)):
\[ i(0) = A + \frac{V_0}{R} \]
Solving for \( A \):
\[ A = i(0) - \frac{V_0}{R} \]
So, the current as a function of time is:
\[ i(t) = \left(i(0) - \frac{V_0}{R}\right) e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
### **Summary:**
- **Transient Response:** \( i(t) = \left(i(0) - \frac{V_0}{R}\right) e^{-\frac{R}{L}t} + \frac{V_0}{R} \)
- **Steady-State Response:** \( i(t) \to \frac{V_0}{R} \) as \( t \to \infty \)
This formula describes how the current changes over time after a voltage is applied to the RL circuit.