How to find current in LR circuit?
2 Answers
To find the current in an LR circuit, which consists of an inductor (L) and a resistor (R) connected in series to a voltage source (V), you can use the principles of electrical circuits and differential equations. Here’s a step-by-step approach to find the current through the circuit:
### 1. **Understand the LR Circuit**
An LR circuit consists of:
- **Inductor (L):** This component opposes changes in current and stores energy in its magnetic field.
- **Resistor (R):** This component opposes the flow of current and dissipates energy as heat.
- **Voltage Source (V):** This provides the electromotive force (EMF) that drives the current through the circuit.
### 2. **Write the Differential Equation**
When a voltage source is applied to an LR circuit, the behavior of the current can be described using Kirchhoff’s Voltage Law (KVL). For a series LR circuit, KVL states that the sum of the voltage drops across the inductor and the resistor equals the voltage of the source.
The voltage across the inductor (\(V_L\)) is given by:
\[ V_L = L \frac{dI(t)}{dt} \]
The voltage across the resistor (\(V_R\)) is given by:
\[ V_R = I(t) R \]
Applying KVL:
\[ V = V_L + V_R \]
\[ V = L \frac{dI(t)}{dt} + I(t) R \]
Rearrange to form a differential equation:
\[ L \frac{dI(t)}{dt} + I(t) R = V \]
### 3. **Solve the Differential Equation**
This is a first-order linear differential equation. To solve it, follow these steps:
#### a. **Find the Homogeneous Solution**
First, solve the homogeneous equation:
\[ L \frac{dI_h(t)}{dt} + I_h(t) R = 0 \]
Rearrange to:
\[ \frac{dI_h(t)}{dt} = -\frac{R}{L} I_h(t) \]
This can be solved as:
\[ I_h(t) = I_0 e^{-\frac{R}{L} t} \]
where \(I_0\) is the initial current at \(t = 0\).
#### b. **Find the Particular Solution**
Next, find a particular solution to the non-homogeneous equation. Since the voltage \(V\) is a constant, assume a constant particular solution \(I_p\). Substituting \(I_p\) into the differential equation:
\[ L \cdot 0 + I_p R = V \]
\[ I_p = \frac{V}{R} \]
#### c. **Combine Solutions**
The total solution is the sum of the homogeneous and particular solutions:
\[ I(t) = I_h(t) + I_p \]
\[ I(t) = I_0 e^{-\frac{R}{L} t} + \frac{V}{R} \]
#### d. **Determine the Initial Condition**
To find \(I_0\), use the initial condition of the circuit. Typically, at \(t = 0\), the inductor has not had time to build up any current (assuming it starts from rest):
\[ I(0) = 0 \]
So:
\[ 0 = I_0 e^{0} + \frac{V}{R} \]
\[ I_0 = -\frac{V}{R} \]
Thus:
\[ I(t) = -\frac{V}{R} e^{-\frac{R}{L} t} + \frac{V}{R} \]
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
### 4. **Current Over Time**
The final expression for the current in the LR circuit as a function of time is:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
This shows that the current starts at 0 and asymptotically approaches \(\frac{V}{R}\) as time progresses, following an exponential growth curve.
### Summary
To find the current in an LR circuit:
1. Write the differential equation based on KVL.
2. Solve the differential equation for the current \(I(t)\).
3. Apply initial conditions to determine constants.
4. Use the solution to find the current at any given time.
This method provides a comprehensive way to understand how the current evolves in an LR circuit over time.
### 1. **Understand the LR Circuit**
An LR circuit consists of:
- **Inductor (L):** This component opposes changes in current and stores energy in its magnetic field.
- **Resistor (R):** This component opposes the flow of current and dissipates energy as heat.
- **Voltage Source (V):** This provides the electromotive force (EMF) that drives the current through the circuit.
### 2. **Write the Differential Equation**
When a voltage source is applied to an LR circuit, the behavior of the current can be described using Kirchhoff’s Voltage Law (KVL). For a series LR circuit, KVL states that the sum of the voltage drops across the inductor and the resistor equals the voltage of the source.
The voltage across the inductor (\(V_L\)) is given by:
\[ V_L = L \frac{dI(t)}{dt} \]
The voltage across the resistor (\(V_R\)) is given by:
\[ V_R = I(t) R \]
Applying KVL:
\[ V = V_L + V_R \]
\[ V = L \frac{dI(t)}{dt} + I(t) R \]
Rearrange to form a differential equation:
\[ L \frac{dI(t)}{dt} + I(t) R = V \]
### 3. **Solve the Differential Equation**
This is a first-order linear differential equation. To solve it, follow these steps:
#### a. **Find the Homogeneous Solution**
First, solve the homogeneous equation:
\[ L \frac{dI_h(t)}{dt} + I_h(t) R = 0 \]
Rearrange to:
\[ \frac{dI_h(t)}{dt} = -\frac{R}{L} I_h(t) \]
This can be solved as:
\[ I_h(t) = I_0 e^{-\frac{R}{L} t} \]
where \(I_0\) is the initial current at \(t = 0\).
#### b. **Find the Particular Solution**
Next, find a particular solution to the non-homogeneous equation. Since the voltage \(V\) is a constant, assume a constant particular solution \(I_p\). Substituting \(I_p\) into the differential equation:
\[ L \cdot 0 + I_p R = V \]
\[ I_p = \frac{V}{R} \]
#### c. **Combine Solutions**
The total solution is the sum of the homogeneous and particular solutions:
\[ I(t) = I_h(t) + I_p \]
\[ I(t) = I_0 e^{-\frac{R}{L} t} + \frac{V}{R} \]
#### d. **Determine the Initial Condition**
To find \(I_0\), use the initial condition of the circuit. Typically, at \(t = 0\), the inductor has not had time to build up any current (assuming it starts from rest):
\[ I(0) = 0 \]
So:
\[ 0 = I_0 e^{0} + \frac{V}{R} \]
\[ I_0 = -\frac{V}{R} \]
Thus:
\[ I(t) = -\frac{V}{R} e^{-\frac{R}{L} t} + \frac{V}{R} \]
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
### 4. **Current Over Time**
The final expression for the current in the LR circuit as a function of time is:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right) \]
This shows that the current starts at 0 and asymptotically approaches \(\frac{V}{R}\) as time progresses, following an exponential growth curve.
### Summary
To find the current in an LR circuit:
1. Write the differential equation based on KVL.
2. Solve the differential equation for the current \(I(t)\).
3. Apply initial conditions to determine constants.
4. Use the solution to find the current at any given time.
This method provides a comprehensive way to understand how the current evolves in an LR circuit over time.
To find the current in an LR circuit, which consists of an inductor (L) and a resistor (R) connected in series, you need to solve for the current over time. The method you use will depend on whether the circuit is in a transient state (e.g., when the circuit is first turned on or off) or in a steady-state (e.g., after a long time has passed).
### 1. **Transient Analysis (When Switching On or Off)**
#### **When the Circuit is Switched On:**
When a DC voltage source \( V \) is applied to an LR circuit that was initially at rest, the circuit will exhibit transient behavior until it reaches a steady state. The differential equation governing the circuit is derived from Kirchhoff's Voltage Law (KVL):
\[ V(t) = L \frac{dI(t)}{dt} + I(t)R \]
Here, \( I(t) \) is the current at time \( t \), \( L \) is the inductance, \( R \) is the resistance, and \( V(t) \) is the applied voltage. For a step input where the voltage suddenly changes to \( V_0 \) at \( t = 0 \), the equation simplifies to:
\[ V_0 = L \frac{dI(t)}{dt} + I(t)R \]
To solve this differential equation, follow these steps:
1. **Find the homogeneous solution:**
Solve the homogeneous equation:
\[ 0 = L \frac{dI_h(t)}{dt} + I_h(t)R \]
The solution is:
\[ I_h(t) = A e^{-\frac{R}{L}t} \]
where \( A \) is a constant determined by initial conditions.
2. **Find the particular solution:**
For a step input \( V_0 \), the particular solution is a constant:
\[ I_p(t) = \frac{V_0}{R} \]
3. **Combine solutions and apply initial conditions:**
The total current \( I(t) \) is:
\[ I(t) = I_h(t) + I_p(t) = A e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
Given that the initial current \( I(0) \) is zero (if the circuit was initially at rest), solve for \( A \):
\[ I(0) = A + \frac{V_0}{R} = 0 \]
\[ A = -\frac{V_0}{R} \]
Thus:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
#### **When the Circuit is Switched Off:**
If the circuit is suddenly disconnected from the voltage source at \( t = 0 \), the initial current is \( I_0 \), and the voltage source is zero. The equation then becomes:
\[ 0 = L \frac{dI(t)}{dt} + I(t)R \]
Solving this yields:
\[ I(t) = I_0 e^{-\frac{R}{L}t} \]
### 2. **Steady-State Analysis (Long-Term Behavior)**
In the steady state (a long time after the switch is closed), the inductor behaves like a short circuit (since \( \frac{dI(t)}{dt} \to 0 \)), and the current \( I \) is:
\[ I_{\text{steady}} = \frac{V}{R} \]
where \( V \) is the DC voltage applied to the circuit.
### Summary
- **Transient Analysis:** Use the differential equation \( V(t) = L \frac{dI(t)}{dt} + I(t)R \) and solve for \( I(t) \) considering initial conditions.
- **Steady-State Analysis:** For long-term behavior, the inductor acts as a short circuit, and the current is \( I = \frac{V}{R} \).
The exact method will depend on whether you are dealing with switching on or off, and whether you need to solve the circuit behavior transiently or just find the final steady-state current.
### 1. **Transient Analysis (When Switching On or Off)**
#### **When the Circuit is Switched On:**
When a DC voltage source \( V \) is applied to an LR circuit that was initially at rest, the circuit will exhibit transient behavior until it reaches a steady state. The differential equation governing the circuit is derived from Kirchhoff's Voltage Law (KVL):
\[ V(t) = L \frac{dI(t)}{dt} + I(t)R \]
Here, \( I(t) \) is the current at time \( t \), \( L \) is the inductance, \( R \) is the resistance, and \( V(t) \) is the applied voltage. For a step input where the voltage suddenly changes to \( V_0 \) at \( t = 0 \), the equation simplifies to:
\[ V_0 = L \frac{dI(t)}{dt} + I(t)R \]
To solve this differential equation, follow these steps:
1. **Find the homogeneous solution:**
Solve the homogeneous equation:
\[ 0 = L \frac{dI_h(t)}{dt} + I_h(t)R \]
The solution is:
\[ I_h(t) = A e^{-\frac{R}{L}t} \]
where \( A \) is a constant determined by initial conditions.
2. **Find the particular solution:**
For a step input \( V_0 \), the particular solution is a constant:
\[ I_p(t) = \frac{V_0}{R} \]
3. **Combine solutions and apply initial conditions:**
The total current \( I(t) \) is:
\[ I(t) = I_h(t) + I_p(t) = A e^{-\frac{R}{L}t} + \frac{V_0}{R} \]
Given that the initial current \( I(0) \) is zero (if the circuit was initially at rest), solve for \( A \):
\[ I(0) = A + \frac{V_0}{R} = 0 \]
\[ A = -\frac{V_0}{R} \]
Thus:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
#### **When the Circuit is Switched Off:**
If the circuit is suddenly disconnected from the voltage source at \( t = 0 \), the initial current is \( I_0 \), and the voltage source is zero. The equation then becomes:
\[ 0 = L \frac{dI(t)}{dt} + I(t)R \]
Solving this yields:
\[ I(t) = I_0 e^{-\frac{R}{L}t} \]
### 2. **Steady-State Analysis (Long-Term Behavior)**
In the steady state (a long time after the switch is closed), the inductor behaves like a short circuit (since \( \frac{dI(t)}{dt} \to 0 \)), and the current \( I \) is:
\[ I_{\text{steady}} = \frac{V}{R} \]
where \( V \) is the DC voltage applied to the circuit.
### Summary
- **Transient Analysis:** Use the differential equation \( V(t) = L \frac{dI(t)}{dt} + I(t)R \) and solve for \( I(t) \) considering initial conditions.
- **Steady-State Analysis:** For long-term behavior, the inductor acts as a short circuit, and the current is \( I = \frac{V}{R} \).
The exact method will depend on whether you are dealing with switching on or off, and whether you need to solve the circuit behavior transiently or just find the final steady-state current.