What is the equation for the LR circuit?
2 Answers
The equation for a series LR circuit, which consists of an inductor (L) and a resistor (R) connected in series with a voltage source (V), is derived from Kirchhoff's voltage law. The fundamental differential equation governing the circuit can be expressed as:
\[
V = L \frac{di(t)}{dt} + Ri(t)
\]
Where:
- \( V \) is the voltage supplied by the source.
- \( i(t) \) is the current at time \( t \).
- \( L \) is the inductance in henries (H).
- \( R \) is the resistance in ohms (Ω).
To solve this equation, we typically rearrange it to:
\[
L \frac{di(t)}{dt} + Ri(t) = V
\]
This is a first-order linear differential equation. The solution will vary depending on whether you're looking for the transient response (how the current changes over time after the switch is closed) or the steady-state response (the final current after a long time).
### Transient Response
For the transient response, if the switch is closed at \( t = 0 \), the current \( i(t) \) can be expressed as:
\[
i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
Where \( \frac{V}{R} \) is the steady-state current, and \( e^{-\frac{R}{L}t} \) represents the transient behavior of the current as it approaches the steady state.
### Steady-State Response
After a long time, the current stabilizes to:
\[
i(t) \to \frac{V}{R}
\]
This indicates that the inductor behaves like a short circuit, and the full voltage drop occurs across the resistor.
If you have any specific aspect you'd like to explore further, feel free to ask!
\[
V = L \frac{di(t)}{dt} + Ri(t)
\]
Where:
- \( V \) is the voltage supplied by the source.
- \( i(t) \) is the current at time \( t \).
- \( L \) is the inductance in henries (H).
- \( R \) is the resistance in ohms (Ω).
To solve this equation, we typically rearrange it to:
\[
L \frac{di(t)}{dt} + Ri(t) = V
\]
This is a first-order linear differential equation. The solution will vary depending on whether you're looking for the transient response (how the current changes over time after the switch is closed) or the steady-state response (the final current after a long time).
### Transient Response
For the transient response, if the switch is closed at \( t = 0 \), the current \( i(t) \) can be expressed as:
\[
i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
Where \( \frac{V}{R} \) is the steady-state current, and \( e^{-\frac{R}{L}t} \) represents the transient behavior of the current as it approaches the steady state.
### Steady-State Response
After a long time, the current stabilizes to:
\[
i(t) \to \frac{V}{R}
\]
This indicates that the inductor behaves like a short circuit, and the full voltage drop occurs across the resistor.
If you have any specific aspect you'd like to explore further, feel free to ask!
An **LR circuit** is a circuit consisting of an inductor (denoted by **L**) and a resistor (denoted by **R**) connected either in series or parallel with a voltage source. The equation for an **LR series circuit** is derived using **Kirchhoff's voltage law (KVL)**, which states that the sum of the voltage drops in a closed loop equals the applied voltage.
### Series LR Circuit Equation
The total voltage **V(t)** applied to the circuit is the sum of the voltage drop across the resistor **V_R(t)** and the voltage drop across the inductor **V_L(t)**. According to Ohm's law and Faraday's law of electromagnetic induction:
- Voltage across the resistor:
\[
V_R(t) = I(t) \cdot R
\]
where:
- \( I(t) \) is the current through the resistor (and also through the inductor, as it's a series circuit).
- \( R \) is the resistance.
- Voltage across the inductor:
\[
V_L(t) = L \cdot \frac{dI(t)}{dt}
\]
where:
- \( L \) is the inductance.
- \( \frac{dI(t)}{dt} \) is the rate of change of current through the inductor.
Using Kirchhoff's voltage law:
\[
V(t) = V_R(t) + V_L(t)
\]
Substituting the expressions for \( V_R(t) \) and \( V_L(t) \):
\[
V(t) = I(t) \cdot R + L \cdot \frac{dI(t)}{dt}
\]
This is the **differential equation** that governs the behavior of an LR series circuit.
### Solving the Equation for Current \( I(t) \)
For a step input (e.g., if a constant voltage \( V_0 \) is applied at time \( t = 0 \)):
\[
V(t) = V_0
\]
The differential equation becomes:
\[
V_0 = I(t) \cdot R + L \cdot \frac{dI(t)}{dt}
\]
To solve this, you can rearrange it as:
\[
\frac{dI(t)}{dt} + \frac{R}{L} I(t) = \frac{V_0}{L}
\]
This is a first-order linear differential equation. The solution for \( I(t) \), the current as a function of time, is:
\[
I(t) = \frac{V_0}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
where:
- \( \frac{V_0}{R} \) is the steady-state current (after a long time),
- \( e^{-\frac{R}{L}t} \) is the exponential decay term,
- \( \frac{R}{L} \) is the time constant \( \tau \), which characterizes the rate at which the current reaches its final value.
### Time Constant
The time constant \( \tau \) for an LR circuit is defined as:
\[
\tau = \frac{L}{R}
\]
This represents the time it takes for the current to reach approximately 63% of its final steady-state value after the voltage is applied.
### Summary of Key Points
- **Differential equation** for an LR circuit:
\[
V(t) = I(t) \cdot R + L \cdot \frac{dI(t)}{dt}
\]
- **Solution for current** \( I(t) \) when a constant voltage \( V_0 \) is applied:
\[
I(t) = \frac{V_0}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
- **Time constant** \( \tau = \frac{L}{R} \), which determines how quickly the current rises or falls.
This equation describes the transient behavior of the current in the circuit as it evolves from zero to its steady-state value.
### Series LR Circuit Equation
The total voltage **V(t)** applied to the circuit is the sum of the voltage drop across the resistor **V_R(t)** and the voltage drop across the inductor **V_L(t)**. According to Ohm's law and Faraday's law of electromagnetic induction:
- Voltage across the resistor:
\[
V_R(t) = I(t) \cdot R
\]
where:
- \( I(t) \) is the current through the resistor (and also through the inductor, as it's a series circuit).
- \( R \) is the resistance.
- Voltage across the inductor:
\[
V_L(t) = L \cdot \frac{dI(t)}{dt}
\]
where:
- \( L \) is the inductance.
- \( \frac{dI(t)}{dt} \) is the rate of change of current through the inductor.
Using Kirchhoff's voltage law:
\[
V(t) = V_R(t) + V_L(t)
\]
Substituting the expressions for \( V_R(t) \) and \( V_L(t) \):
\[
V(t) = I(t) \cdot R + L \cdot \frac{dI(t)}{dt}
\]
This is the **differential equation** that governs the behavior of an LR series circuit.
### Solving the Equation for Current \( I(t) \)
For a step input (e.g., if a constant voltage \( V_0 \) is applied at time \( t = 0 \)):
\[
V(t) = V_0
\]
The differential equation becomes:
\[
V_0 = I(t) \cdot R + L \cdot \frac{dI(t)}{dt}
\]
To solve this, you can rearrange it as:
\[
\frac{dI(t)}{dt} + \frac{R}{L} I(t) = \frac{V_0}{L}
\]
This is a first-order linear differential equation. The solution for \( I(t) \), the current as a function of time, is:
\[
I(t) = \frac{V_0}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
where:
- \( \frac{V_0}{R} \) is the steady-state current (after a long time),
- \( e^{-\frac{R}{L}t} \) is the exponential decay term,
- \( \frac{R}{L} \) is the time constant \( \tau \), which characterizes the rate at which the current reaches its final value.
### Time Constant
The time constant \( \tau \) for an LR circuit is defined as:
\[
\tau = \frac{L}{R}
\]
This represents the time it takes for the current to reach approximately 63% of its final steady-state value after the voltage is applied.
### Summary of Key Points
- **Differential equation** for an LR circuit:
\[
V(t) = I(t) \cdot R + L \cdot \frac{dI(t)}{dt}
\]
- **Solution for current** \( I(t) \) when a constant voltage \( V_0 \) is applied:
\[
I(t) = \frac{V_0}{R} \left( 1 - e^{-\frac{R}{L}t} \right)
\]
- **Time constant** \( \tau = \frac{L}{R} \), which determines how quickly the current rises or falls.
This equation describes the transient behavior of the current in the circuit as it evolves from zero to its steady-state value.