What is the differential equation for the LR series circuit?
2 Answers
In an LR series circuit, where an inductor (L) and a resistor (R) are connected in series, the differential equation describing the circuit behavior is derived from Kirchhoff’s voltage law (KVL).
Here’s how you can derive it:
1. **Kirchhoff’s Voltage Law (KVL):** The sum of the voltages around a closed loop in a circuit must equal zero. For an LR series circuit, this gives:
\[
V(t) = V_R(t) + V_L(t)
\]
where \( V(t) \) is the applied voltage source, \( V_R(t) \) is the voltage across the resistor, and \( V_L(t) \) is the voltage across the inductor.
2. **Voltage across the Resistor (Ohm's Law):**
\[
V_R(t) = i(t) \cdot R
\]
where \( i(t) \) is the current through the resistor.
3. **Voltage across the Inductor:**
The voltage across an inductor is given by:
\[
V_L(t) = L \frac{di(t)}{dt}
\]
4. **Substitute into KVL Equation:**
Combining these, we get:
\[
V(t) = R \cdot i(t) + L \frac{di(t)}{dt}
\]
5. **Rearrange to Form the Differential Equation:**
Rearranging the terms to form a standard differential equation, we have:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V(t)
\]
This is the first-order linear differential equation that governs the current \( i(t) \) in an LR series circuit.
### Summary
The differential equation for an LR series circuit is:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V(t)
\]
where:
- \( L \) is the inductance of the inductor,
- \( R \) is the resistance of the resistor,
- \( i(t) \) is the current through the circuit at time \( t \),
- \( V(t) \) is the applied voltage source at time \( t \).
Here’s how you can derive it:
1. **Kirchhoff’s Voltage Law (KVL):** The sum of the voltages around a closed loop in a circuit must equal zero. For an LR series circuit, this gives:
\[
V(t) = V_R(t) + V_L(t)
\]
where \( V(t) \) is the applied voltage source, \( V_R(t) \) is the voltage across the resistor, and \( V_L(t) \) is the voltage across the inductor.
2. **Voltage across the Resistor (Ohm's Law):**
\[
V_R(t) = i(t) \cdot R
\]
where \( i(t) \) is the current through the resistor.
3. **Voltage across the Inductor:**
The voltage across an inductor is given by:
\[
V_L(t) = L \frac{di(t)}{dt}
\]
4. **Substitute into KVL Equation:**
Combining these, we get:
\[
V(t) = R \cdot i(t) + L \frac{di(t)}{dt}
\]
5. **Rearrange to Form the Differential Equation:**
Rearranging the terms to form a standard differential equation, we have:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V(t)
\]
This is the first-order linear differential equation that governs the current \( i(t) \) in an LR series circuit.
### Summary
The differential equation for an LR series circuit is:
\[
L \frac{di(t)}{dt} + R \cdot i(t) = V(t)
\]
where:
- \( L \) is the inductance of the inductor,
- \( R \) is the resistance of the resistor,
- \( i(t) \) is the current through the circuit at time \( t \),
- \( V(t) \) is the applied voltage source at time \( t \).
An LR series circuit consists of an inductor (\(L\)) and a resistor (\(R\)) connected in series with a voltage source (\(V(t)\)). To derive the differential equation for this circuit, we'll follow these steps:
### 1. Write Down the Voltage and Current Relationships
**Ohm's Law** for the resistor:
\[ V_R = I(t) \cdot R \]
**Faraday's Law** for the inductor:
\[ V_L = L \frac{dI(t)}{dt} \]
In a series circuit, the total voltage across the circuit is the sum of the voltages across the resistor and the inductor:
\[ V(t) = V_R + V_L \]
Substitute the expressions for \( V_R \) and \( V_L \):
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
### 2. Form the Differential Equation
Rearrange the equation to get a standard differential equation form:
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
To isolate the derivative term, rearrange the equation:
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \]
### 3. Express the Differential Equation
This is the differential equation for the current \(I(t)\) in an LR series circuit:
\[ L \frac{dI(t)}{dt} + R \cdot I(t) = V(t) \]
### Summary
In summary, the differential equation governing the current in an LR series circuit, where \(L\) is the inductance, \(R\) is the resistance, and \(V(t)\) is the applied voltage, is:
\[ L \frac{dI(t)}{dt} + R \cdot I(t) = V(t) \]
This equation relates the time derivative of the current through the inductor and the resistor to the applied voltage. Solving this differential equation can help determine how the current \(I(t)\) changes over time in response to different voltage sources \(V(t)\).
### 1. Write Down the Voltage and Current Relationships
**Ohm's Law** for the resistor:
\[ V_R = I(t) \cdot R \]
**Faraday's Law** for the inductor:
\[ V_L = L \frac{dI(t)}{dt} \]
In a series circuit, the total voltage across the circuit is the sum of the voltages across the resistor and the inductor:
\[ V(t) = V_R + V_L \]
Substitute the expressions for \( V_R \) and \( V_L \):
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
### 2. Form the Differential Equation
Rearrange the equation to get a standard differential equation form:
\[ V(t) = I(t) \cdot R + L \frac{dI(t)}{dt} \]
To isolate the derivative term, rearrange the equation:
\[ L \frac{dI(t)}{dt} + I(t) \cdot R = V(t) \]
### 3. Express the Differential Equation
This is the differential equation for the current \(I(t)\) in an LR series circuit:
\[ L \frac{dI(t)}{dt} + R \cdot I(t) = V(t) \]
### Summary
In summary, the differential equation governing the current in an LR series circuit, where \(L\) is the inductance, \(R\) is the resistance, and \(V(t)\) is the applied voltage, is:
\[ L \frac{dI(t)}{dt} + R \cdot I(t) = V(t) \]
This equation relates the time derivative of the current through the inductor and the resistor to the applied voltage. Solving this differential equation can help determine how the current \(I(t)\) changes over time in response to different voltage sources \(V(t)\).