For zero initial conditions, write the equation for current in a series R-L circuit for step response.
2 Answers
In a series R-L circuit with zero initial conditions, the equation for the current \( I(t) \) in response to a step input voltage \( V_0 \) can be derived from Kirchhoff's voltage law. The differential equation governing the circuit is:
\[
V_0 = L \frac{dI(t)}{dt} + RI(t)
\]
This can be rearranged to:
\[
\frac{dI(t)}{dt} + \frac{R}{L} I(t) = \frac{V_0}{L}
\]
To solve this first-order linear differential equation, we can use an integrating factor. The solution is given by:
\[
I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
Here, \( I(t) \) approaches \( \frac{V_0}{R} \) as \( t \) approaches infinity, representing the ultimate steady-state current in the circuit.
\[
V_0 = L \frac{dI(t)}{dt} + RI(t)
\]
This can be rearranged to:
\[
\frac{dI(t)}{dt} + \frac{R}{L} I(t) = \frac{V_0}{L}
\]
To solve this first-order linear differential equation, we can use an integrating factor. The solution is given by:
\[
I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
Here, \( I(t) \) approaches \( \frac{V_0}{R} \) as \( t \) approaches infinity, representing the ultimate steady-state current in the circuit.
Certainly! For a series \( R-L \) circuit with zero initial conditions and a step input, the behavior of the circuit can be described using differential equations. Here’s a step-by-step approach to derive the equation for the current:
### 1. **Circuit Description:**
A series \( R-L \) circuit consists of a resistor \( R \) and an inductor \( L \) connected in series. When a voltage source \( V(t) \) is applied, the circuit will respond according to its differential equation.
### 2. **Step Input Voltage:**
Let's assume a step input voltage \( V(t) \) is applied. This means the voltage source switches from 0V to a constant value \( V_0 \) at \( t = 0 \). Mathematically, this can be expressed as:
\[ V(t) = \begin{cases}
0 & \text{for } t < 0 \\
V_0 & \text{for } t \ge 0
\end{cases} \]
### 3. **Circuit Analysis:**
According to Kirchhoff's Voltage Law (KVL), the sum of the voltages around the loop must equal zero. For a series \( R-L \) circuit, this can be written as:
\[ V(t) = V_R(t) + V_L(t) \]
where \( V_R(t) \) is the voltage across the resistor and \( V_L(t) \) is the voltage across the inductor.
The voltage across the resistor \( R \) is given by Ohm’s law:
\[ V_R(t) = i(t)R \]
The voltage across the inductor \( L \) is given by:
\[ V_L(t) = L \frac{di(t)}{dt} \]
Substituting these into KVL:
\[ V(t) = i(t)R + L \frac{di(t)}{dt} \]
### 4. **Differential Equation:**
Substituting the step input \( V(t) = V_0 \) for \( t \ge 0 \):
\[ V_0 = i(t)R + L \frac{di(t)}{dt} \]
Rearranging the equation to isolate the differential term:
\[ L \frac{di(t)}{dt} + Ri(t) = V_0 \]
This is the first-order linear differential equation for the current \( i(t) \).
### 5. **Solving the Differential Equation:**
To solve this, we first find the homogeneous solution and then the particular solution.
**Homogeneous Solution:**
For \( V_0 = 0 \):
\[ L \frac{di(t)}{dt} + Ri(t) = 0 \]
The solution to this homogeneous equation is:
\[ i_h(t) = A e^{-\frac{R}{L} t} \]
where \( A \) is a constant determined by initial conditions.
**Particular Solution:**
For the non-homogeneous part, assume a constant solution \( i_p(t) = I_{\text{steady}} \). Substituting into the differential equation:
\[ L \cdot 0 + R I_{\text{steady}} = V_0 \]
Thus:
\[ I_{\text{steady}} = \frac{V_0}{R} \]
**General Solution:**
Combining the homogeneous and particular solutions:
\[ i(t) = i_h(t) + i_p(t) \]
\[ i(t) = A e^{-\frac{R}{L} t} + \frac{V_0}{R} \]
**Applying Initial Conditions:**
Since the initial current is zero (zero initial conditions), at \( t = 0 \):
\[ i(0) = 0 = A e^{0} + \frac{V_0}{R} \]
\[ 0 = A + \frac{V_0}{R} \]
Thus:
\[ A = -\frac{V_0}{R} \]
Substitute \( A \) back into the general solution:
\[ i(t) = -\frac{V_0}{R} e^{-\frac{R}{L} t} + \frac{V_0}{R} \]
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
### 6. **Final Expression:**
Thus, the current \( i(t) \) in the series \( R-L \) circuit for \( t \ge 0 \) with zero initial conditions is:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
This equation describes how the current increases from zero to its steady-state value \( \frac{V_0}{R} \) over time, governed by the time constant \( \tau = \frac{L}{R} \).
### 1. **Circuit Description:**
A series \( R-L \) circuit consists of a resistor \( R \) and an inductor \( L \) connected in series. When a voltage source \( V(t) \) is applied, the circuit will respond according to its differential equation.
### 2. **Step Input Voltage:**
Let's assume a step input voltage \( V(t) \) is applied. This means the voltage source switches from 0V to a constant value \( V_0 \) at \( t = 0 \). Mathematically, this can be expressed as:
\[ V(t) = \begin{cases}
0 & \text{for } t < 0 \\
V_0 & \text{for } t \ge 0
\end{cases} \]
### 3. **Circuit Analysis:**
According to Kirchhoff's Voltage Law (KVL), the sum of the voltages around the loop must equal zero. For a series \( R-L \) circuit, this can be written as:
\[ V(t) = V_R(t) + V_L(t) \]
where \( V_R(t) \) is the voltage across the resistor and \( V_L(t) \) is the voltage across the inductor.
The voltage across the resistor \( R \) is given by Ohm’s law:
\[ V_R(t) = i(t)R \]
The voltage across the inductor \( L \) is given by:
\[ V_L(t) = L \frac{di(t)}{dt} \]
Substituting these into KVL:
\[ V(t) = i(t)R + L \frac{di(t)}{dt} \]
### 4. **Differential Equation:**
Substituting the step input \( V(t) = V_0 \) for \( t \ge 0 \):
\[ V_0 = i(t)R + L \frac{di(t)}{dt} \]
Rearranging the equation to isolate the differential term:
\[ L \frac{di(t)}{dt} + Ri(t) = V_0 \]
This is the first-order linear differential equation for the current \( i(t) \).
### 5. **Solving the Differential Equation:**
To solve this, we first find the homogeneous solution and then the particular solution.
**Homogeneous Solution:**
For \( V_0 = 0 \):
\[ L \frac{di(t)}{dt} + Ri(t) = 0 \]
The solution to this homogeneous equation is:
\[ i_h(t) = A e^{-\frac{R}{L} t} \]
where \( A \) is a constant determined by initial conditions.
**Particular Solution:**
For the non-homogeneous part, assume a constant solution \( i_p(t) = I_{\text{steady}} \). Substituting into the differential equation:
\[ L \cdot 0 + R I_{\text{steady}} = V_0 \]
Thus:
\[ I_{\text{steady}} = \frac{V_0}{R} \]
**General Solution:**
Combining the homogeneous and particular solutions:
\[ i(t) = i_h(t) + i_p(t) \]
\[ i(t) = A e^{-\frac{R}{L} t} + \frac{V_0}{R} \]
**Applying Initial Conditions:**
Since the initial current is zero (zero initial conditions), at \( t = 0 \):
\[ i(0) = 0 = A e^{0} + \frac{V_0}{R} \]
\[ 0 = A + \frac{V_0}{R} \]
Thus:
\[ A = -\frac{V_0}{R} \]
Substitute \( A \) back into the general solution:
\[ i(t) = -\frac{V_0}{R} e^{-\frac{R}{L} t} + \frac{V_0}{R} \]
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
### 6. **Final Expression:**
Thus, the current \( i(t) \) in the series \( R-L \) circuit for \( t \ge 0 \) with zero initial conditions is:
\[ i(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
This equation describes how the current increases from zero to its steady-state value \( \frac{V_0}{R} \) over time, governed by the time constant \( \tau = \frac{L}{R} \).