How do you find the expression of current in an RL circuit?
2 Answers
To find the expression of current in an RL circuit (resistor-inductor circuit), we need to analyze how the current behaves when the circuit is either charging (after a switch is closed) or discharging (after a switch is opened) in response to the application or removal of a voltage source.
### Step-by-Step Derivation of Current in an RL Circuit (Charging)
Let’s assume an RL circuit consisting of:
- A resistor \( R \)
- An inductor \( L \)
- A DC voltage source \( V \)
- A switch that closes at \( t = 0 \).
When the switch is closed, current begins to flow through the resistor and the inductor. The key here is that an inductor opposes changes in current, causing the current to rise gradually instead of instantaneously.
#### 1. **Kirchhoff’s Voltage Law (KVL):**
When the switch is closed, we can apply KVL to the loop. According to KVL, the sum of the voltage drops across the resistor and the inductor equals the supply voltage \( V \):
\[
V = V_R + V_L
\]
Where:
- \( V_R \) is the voltage drop across the resistor, and according to Ohm’s law, \( V_R = i(t)R \).
- \( V_L \) is the voltage across the inductor, and the voltage across an inductor is given by \( V_L = L \frac{di(t)}{dt} \), where \( \frac{di(t)}{dt} \) is the rate of change of current.
Substituting these into the KVL equation:
\[
V = i(t)R + L \frac{di(t)}{dt}
\]
#### 2. **Rearranging the Differential Equation:**
We want to solve this equation for \( i(t) \), the current as a function of time.
\[
L \frac{di(t)}{dt} + i(t)R = V
\]
This is a first-order linear differential equation. To solve it, we use standard techniques for solving such equations.
#### 3. **Solving the Differential Equation:**
First, we rearrange the equation:
\[
\frac{di(t)}{dt} + \frac{R}{L}i(t) = \frac{V}{L}
\]
Now, let’s solve this using an integrating factor.
#### 4. **Using the Integrating Factor:**
The general form of a first-order linear differential equation is:
\[
\frac{dy}{dt} + P(t)y = Q(t)
\]
In this case, \( P(t) = \frac{R}{L} \) and \( Q(t) = \frac{V}{L} \).
The integrating factor is given by:
\[
\mu(t) = e^{\int P(t) dt} = e^{\int \frac{R}{L} dt} = e^{\frac{R}{L}t}
\]
We multiply both sides of the differential equation by this integrating factor:
\[
e^{\frac{R}{L}t} \frac{di(t)}{dt} + \frac{R}{L} e^{\frac{R}{L}t} i(t) = \frac{V}{L} e^{\frac{R}{L}t}
\]
The left-hand side is the derivative of \( e^{\frac{R}{L}t} i(t) \), so we can rewrite it as:
\[
\frac{d}{dt} \left( e^{\frac{R}{L}t} i(t) \right) = \frac{V}{L} e^{\frac{R}{L}t}
\]
#### 5. **Integrating Both Sides:**
Integrating both sides with respect to time:
\[
e^{\frac{R}{L}t} i(t) = \int \frac{V}{L} e^{\frac{R}{L}t} dt
\]
The integral of \( e^{\frac{R}{L}t} \) is \( \frac{L}{R} e^{\frac{R}{L}t} \), so we have:
\[
e^{\frac{R}{L}t} i(t) = \frac{V}{R} e^{\frac{R}{L}t} + C
\]
Where \( C \) is the constant of integration. To solve for \( i(t) \), we divide both sides by \( e^{\frac{R}{L}t} \):
\[
i(t) = \frac{V}{R} + Ce^{-\frac{R}{L}t}
\]
#### 6. **Applying Initial Conditions:**
At \( t = 0 \), the initial current in the circuit is typically zero (since the inductor resists any instantaneous change in current). So,
\[
i(0) = 0
\]
Substitute \( t = 0 \) into the equation:
\[
0 = \frac{V}{R} + C e^{0}
\]
\[
0 = \frac{V}{R} + C
\]
\[
C = -\frac{V}{R}
\]
Thus, the expression for the current becomes:
\[
i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
### Final Expression for Current in an RL Circuit:
The current in the RL circuit as a function of time, for the charging phase, is:
\[
i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
This shows that the current starts at zero when \( t = 0 \) and asymptotically approaches \( \frac{V}{R} \) as \( t \to \infty \). The time constant \( \tau = \frac{L}{R} \) characterizes how fast the current rises. After a time of approximately \( 5\tau \), the current reaches over 99% of its final value.
### Current in an RL Circuit (Discharging):
If the voltage source is removed (i.e., the switch is opened), the inductor will try to maintain the current, but the current will decay over time.
For the discharging case, the KVL equation becomes:
\[
0 = i(t)R + L \frac{di(t)}{dt}
\]
Following a similar process as before, the solution is:
\[
i(t) = I_0 e^{-\frac{R}{L}t}
\]
Where \( I_0 \) is the initial current at \( t = 0 \) (just before the switch is opened).
### Summary:
- **Charging:** \( i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right) \)
- **Discharging:** \( i(t) = I_0 e^{-\frac{R}{L}t} \)
These expressions describe the behavior of the current in an RL circuit over time, depending on whether the circuit is charging or discharging.
### Step-by-Step Derivation of Current in an RL Circuit (Charging)
Let’s assume an RL circuit consisting of:
- A resistor \( R \)
- An inductor \( L \)
- A DC voltage source \( V \)
- A switch that closes at \( t = 0 \).
When the switch is closed, current begins to flow through the resistor and the inductor. The key here is that an inductor opposes changes in current, causing the current to rise gradually instead of instantaneously.
#### 1. **Kirchhoff’s Voltage Law (KVL):**
When the switch is closed, we can apply KVL to the loop. According to KVL, the sum of the voltage drops across the resistor and the inductor equals the supply voltage \( V \):
\[
V = V_R + V_L
\]
Where:
- \( V_R \) is the voltage drop across the resistor, and according to Ohm’s law, \( V_R = i(t)R \).
- \( V_L \) is the voltage across the inductor, and the voltage across an inductor is given by \( V_L = L \frac{di(t)}{dt} \), where \( \frac{di(t)}{dt} \) is the rate of change of current.
Substituting these into the KVL equation:
\[
V = i(t)R + L \frac{di(t)}{dt}
\]
#### 2. **Rearranging the Differential Equation:**
We want to solve this equation for \( i(t) \), the current as a function of time.
\[
L \frac{di(t)}{dt} + i(t)R = V
\]
This is a first-order linear differential equation. To solve it, we use standard techniques for solving such equations.
#### 3. **Solving the Differential Equation:**
First, we rearrange the equation:
\[
\frac{di(t)}{dt} + \frac{R}{L}i(t) = \frac{V}{L}
\]
Now, let’s solve this using an integrating factor.
#### 4. **Using the Integrating Factor:**
The general form of a first-order linear differential equation is:
\[
\frac{dy}{dt} + P(t)y = Q(t)
\]
In this case, \( P(t) = \frac{R}{L} \) and \( Q(t) = \frac{V}{L} \).
The integrating factor is given by:
\[
\mu(t) = e^{\int P(t) dt} = e^{\int \frac{R}{L} dt} = e^{\frac{R}{L}t}
\]
We multiply both sides of the differential equation by this integrating factor:
\[
e^{\frac{R}{L}t} \frac{di(t)}{dt} + \frac{R}{L} e^{\frac{R}{L}t} i(t) = \frac{V}{L} e^{\frac{R}{L}t}
\]
The left-hand side is the derivative of \( e^{\frac{R}{L}t} i(t) \), so we can rewrite it as:
\[
\frac{d}{dt} \left( e^{\frac{R}{L}t} i(t) \right) = \frac{V}{L} e^{\frac{R}{L}t}
\]
#### 5. **Integrating Both Sides:**
Integrating both sides with respect to time:
\[
e^{\frac{R}{L}t} i(t) = \int \frac{V}{L} e^{\frac{R}{L}t} dt
\]
The integral of \( e^{\frac{R}{L}t} \) is \( \frac{L}{R} e^{\frac{R}{L}t} \), so we have:
\[
e^{\frac{R}{L}t} i(t) = \frac{V}{R} e^{\frac{R}{L}t} + C
\]
Where \( C \) is the constant of integration. To solve for \( i(t) \), we divide both sides by \( e^{\frac{R}{L}t} \):
\[
i(t) = \frac{V}{R} + Ce^{-\frac{R}{L}t}
\]
#### 6. **Applying Initial Conditions:**
At \( t = 0 \), the initial current in the circuit is typically zero (since the inductor resists any instantaneous change in current). So,
\[
i(0) = 0
\]
Substitute \( t = 0 \) into the equation:
\[
0 = \frac{V}{R} + C e^{0}
\]
\[
0 = \frac{V}{R} + C
\]
\[
C = -\frac{V}{R}
\]
Thus, the expression for the current becomes:
\[
i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
### Final Expression for Current in an RL Circuit:
The current in the RL circuit as a function of time, for the charging phase, is:
\[
i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
This shows that the current starts at zero when \( t = 0 \) and asymptotically approaches \( \frac{V}{R} \) as \( t \to \infty \). The time constant \( \tau = \frac{L}{R} \) characterizes how fast the current rises. After a time of approximately \( 5\tau \), the current reaches over 99% of its final value.
### Current in an RL Circuit (Discharging):
If the voltage source is removed (i.e., the switch is opened), the inductor will try to maintain the current, but the current will decay over time.
For the discharging case, the KVL equation becomes:
\[
0 = i(t)R + L \frac{di(t)}{dt}
\]
Following a similar process as before, the solution is:
\[
i(t) = I_0 e^{-\frac{R}{L}t}
\]
Where \( I_0 \) is the initial current at \( t = 0 \) (just before the switch is opened).
### Summary:
- **Charging:** \( i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right) \)
- **Discharging:** \( i(t) = I_0 e^{-\frac{R}{L}t} \)
These expressions describe the behavior of the current in an RL circuit over time, depending on whether the circuit is charging or discharging.
To find the expression for the current in an RL circuit, you need to consider the nature of the circuit, which typically consists of a resistor (R) and an inductor (L) in series. The analysis varies depending on whether the circuit is driven by a constant voltage source or an alternating current (AC) source. Below are the methods for each case:
### 1. **RL Circuit with a DC Source**
**Given:**
- A series circuit with a resistor \( R \) and an inductor \( L \).
- A constant voltage source \( V \) applied at \( t = 0 \).
**Objective:**
Find the current \( I(t) \) as a function of time.
**Solution:**
1. **Write the Differential Equation:**
Applying Kirchhoff’s Voltage Law (KVL) around the circuit, we get:
\[
V = V_R + V_L
\]
where \( V_R \) is the voltage across the resistor and \( V_L \) is the voltage across the inductor.
For the resistor:
\[
V_R = I(t) \cdot R
\]
For the inductor:
\[
V_L = L \frac{dI(t)}{dt}
\]
Combining these:
\[
V = I(t) \cdot R + L \frac{dI(t)}{dt}
\]
2. **Solve the Differential Equation:**
Rearrange to form a standard first-order linear differential equation:
\[
L \frac{dI(t)}{dt} + R I(t) = V
\]
The solution involves finding the homogeneous and particular solutions.
- **Homogeneous Solution:**
Solve the homogeneous differential equation:
\[
L \frac{dI_h(t)}{dt} + R I_h(t) = 0
\]
which simplifies to:
\[
\frac{dI_h(t)}{dt} = -\frac{R}{L} I_h(t)
\]
The solution is:
\[
I_h(t) = A e^{-\frac{R}{L} t}
\]
- **Particular Solution:**
For a constant voltage \( V \), the particular solution is:
\[
I_p(t) = \frac{V}{R}
\]
- **Combine Solutions:**
The general solution is:
\[
I(t) = I_h(t) + I_p(t) = A e^{-\frac{R}{L} t} + \frac{V}{R}
\]
- **Determine the Constant \( A \):**
Use the initial condition \( I(0) = 0 \) (assuming the circuit is at rest before \( t = 0 \)):
\[
0 = A e^{0} + \frac{V}{R}
\]
Hence:
\[
A = -\frac{V}{R}
\]
- **Final Expression:**
Therefore, the current in the circuit is:
\[
I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right)
\]
### 2. **RL Circuit with an AC Source**
**Given:**
- A series circuit with a resistor \( R \) and an inductor \( L \).
- An alternating current source with voltage \( V(t) = V_0 \sin(\omega t) \).
**Objective:**
Find the current \( I(t) \).
**Solution:**
1. **Write the Differential Equation:**
Using KVL, the voltage source and the components' voltages:
\[
V(t) = V_R + V_L
\]
which translates to:
\[
V_0 \sin(\omega t) = I(t) R + L \frac{dI(t)}{dt}
\]
2. **Solve Using Phasors:**
To solve this using phasors, assume a sinusoidal steady-state response:
\[
I(t) = I_0 \sin(\omega t + \phi)
\]
Convert the circuit to its phasor equivalent:
\[
V_0 = I_0 (R + j\omega L)
\]
where \( j \) is the imaginary unit.
The magnitude of the impedance \( Z \) is:
\[
Z = \sqrt{R^2 + (\omega L)^2}
\]
The phasor current is:
\[
I_0 = \frac{V_0}{Z}
\]
Therefore:
\[
I(t) = \frac{V_0}{\sqrt{R^2 + (\omega L)^2}} \sin(\omega t + \phi)
\]
where \( \phi \) is the phase angle given by:
\[
\phi = \arctan \left(-\frac{\omega L}{R}\right)
\]
### Summary:
- For a DC source, the current \( I(t) \) starts from zero and asymptotically approaches \( \frac{V}{R} \) as \( t \) goes to infinity.
- For an AC source, the current is sinusoidal, with its amplitude and phase depending on the resistance and inductance of the circuit.
These methods cover the most common scenarios for finding the current in an RL circuit. If you have a specific situation or different source type, let me know!
### 1. **RL Circuit with a DC Source**
**Given:**
- A series circuit with a resistor \( R \) and an inductor \( L \).
- A constant voltage source \( V \) applied at \( t = 0 \).
**Objective:**
Find the current \( I(t) \) as a function of time.
**Solution:**
1. **Write the Differential Equation:**
Applying Kirchhoff’s Voltage Law (KVL) around the circuit, we get:
\[
V = V_R + V_L
\]
where \( V_R \) is the voltage across the resistor and \( V_L \) is the voltage across the inductor.
For the resistor:
\[
V_R = I(t) \cdot R
\]
For the inductor:
\[
V_L = L \frac{dI(t)}{dt}
\]
Combining these:
\[
V = I(t) \cdot R + L \frac{dI(t)}{dt}
\]
2. **Solve the Differential Equation:**
Rearrange to form a standard first-order linear differential equation:
\[
L \frac{dI(t)}{dt} + R I(t) = V
\]
The solution involves finding the homogeneous and particular solutions.
- **Homogeneous Solution:**
Solve the homogeneous differential equation:
\[
L \frac{dI_h(t)}{dt} + R I_h(t) = 0
\]
which simplifies to:
\[
\frac{dI_h(t)}{dt} = -\frac{R}{L} I_h(t)
\]
The solution is:
\[
I_h(t) = A e^{-\frac{R}{L} t}
\]
- **Particular Solution:**
For a constant voltage \( V \), the particular solution is:
\[
I_p(t) = \frac{V}{R}
\]
- **Combine Solutions:**
The general solution is:
\[
I(t) = I_h(t) + I_p(t) = A e^{-\frac{R}{L} t} + \frac{V}{R}
\]
- **Determine the Constant \( A \):**
Use the initial condition \( I(0) = 0 \) (assuming the circuit is at rest before \( t = 0 \)):
\[
0 = A e^{0} + \frac{V}{R}
\]
Hence:
\[
A = -\frac{V}{R}
\]
- **Final Expression:**
Therefore, the current in the circuit is:
\[
I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right)
\]
### 2. **RL Circuit with an AC Source**
**Given:**
- A series circuit with a resistor \( R \) and an inductor \( L \).
- An alternating current source with voltage \( V(t) = V_0 \sin(\omega t) \).
**Objective:**
Find the current \( I(t) \).
**Solution:**
1. **Write the Differential Equation:**
Using KVL, the voltage source and the components' voltages:
\[
V(t) = V_R + V_L
\]
which translates to:
\[
V_0 \sin(\omega t) = I(t) R + L \frac{dI(t)}{dt}
\]
2. **Solve Using Phasors:**
To solve this using phasors, assume a sinusoidal steady-state response:
\[
I(t) = I_0 \sin(\omega t + \phi)
\]
Convert the circuit to its phasor equivalent:
\[
V_0 = I_0 (R + j\omega L)
\]
where \( j \) is the imaginary unit.
The magnitude of the impedance \( Z \) is:
\[
Z = \sqrt{R^2 + (\omega L)^2}
\]
The phasor current is:
\[
I_0 = \frac{V_0}{Z}
\]
Therefore:
\[
I(t) = \frac{V_0}{\sqrt{R^2 + (\omega L)^2}} \sin(\omega t + \phi)
\]
where \( \phi \) is the phase angle given by:
\[
\phi = \arctan \left(-\frac{\omega L}{R}\right)
\]
### Summary:
- For a DC source, the current \( I(t) \) starts from zero and asymptotically approaches \( \frac{V}{R} \) as \( t \) goes to infinity.
- For an AC source, the current is sinusoidal, with its amplitude and phase depending on the resistance and inductance of the circuit.
These methods cover the most common scenarios for finding the current in an RL circuit. If you have a specific situation or different source type, let me know!