What is the expression for the decay of current in RL circuit?
2 Answers
In an RL circuit (a circuit with a resistor \( R \) and an inductor \( L \) in series), the decay of current over time when the circuit is suddenly disconnected from a voltage source can be described by the following expression:
\[ I(t) = I_0 \cdot e^{-\frac{R}{L} t} \]
where:
- \( I(t) \) is the current through the circuit at time \( t \).
- \( I_0 \) is the initial current at \( t = 0 \).
- \( R \) is the resistance in ohms (Ω).
- \( L \) is the inductance in henries (H).
- \( t \) is the time in seconds (s).
### Derivation
1. **Kirchhoff's Voltage Law (KVL):**
When the circuit is disconnected from the power source, the only components present are the resistor \( R \) and the inductor \( L \). According to KVL, the sum of the voltage drops across these components must equal zero:
\[ V_R + V_L = 0 \]
2. **Voltage across the Resistor (Ohm's Law):**
The voltage drop across the resistor is:
\[ V_R = I(t) \cdot R \]
3. **Voltage across the Inductor:**
The voltage drop across the inductor is given by:
\[ V_L = L \frac{dI(t)}{dt} \]
4. **Setting up the Differential Equation:**
Substituting these into the KVL equation:
\[ I(t) \cdot R + L \frac{dI(t)}{dt} = 0 \]
5. **Solving the Differential Equation:**
Rearranging gives:
\[ \frac{dI(t)}{dt} = -\frac{R}{L} I(t) \]
This is a first-order linear differential equation. Its solution is of the form:
\[ I(t) = I_0 \cdot e^{-\frac{R}{L} t} \]
where \( I_0 \) is the current at \( t = 0 \), which is the initial current flowing through the circuit just before the switch was opened.
### Summary
The expression \( I(t) = I_0 \cdot e^{-\frac{R}{L} t} \) represents the exponential decay of current in an RL circuit when the circuit is disconnected from a voltage source, where the rate of decay is determined by the time constant \( \tau = \frac{L}{R} \).
\[ I(t) = I_0 \cdot e^{-\frac{R}{L} t} \]
where:
- \( I(t) \) is the current through the circuit at time \( t \).
- \( I_0 \) is the initial current at \( t = 0 \).
- \( R \) is the resistance in ohms (Ω).
- \( L \) is the inductance in henries (H).
- \( t \) is the time in seconds (s).
### Derivation
1. **Kirchhoff's Voltage Law (KVL):**
When the circuit is disconnected from the power source, the only components present are the resistor \( R \) and the inductor \( L \). According to KVL, the sum of the voltage drops across these components must equal zero:
\[ V_R + V_L = 0 \]
2. **Voltage across the Resistor (Ohm's Law):**
The voltage drop across the resistor is:
\[ V_R = I(t) \cdot R \]
3. **Voltage across the Inductor:**
The voltage drop across the inductor is given by:
\[ V_L = L \frac{dI(t)}{dt} \]
4. **Setting up the Differential Equation:**
Substituting these into the KVL equation:
\[ I(t) \cdot R + L \frac{dI(t)}{dt} = 0 \]
5. **Solving the Differential Equation:**
Rearranging gives:
\[ \frac{dI(t)}{dt} = -\frac{R}{L} I(t) \]
This is a first-order linear differential equation. Its solution is of the form:
\[ I(t) = I_0 \cdot e^{-\frac{R}{L} t} \]
where \( I_0 \) is the current at \( t = 0 \), which is the initial current flowing through the circuit just before the switch was opened.
### Summary
The expression \( I(t) = I_0 \cdot e^{-\frac{R}{L} t} \) represents the exponential decay of current in an RL circuit when the circuit is disconnected from a voltage source, where the rate of decay is determined by the time constant \( \tau = \frac{L}{R} \).
In an RL circuit, which consists of a resistor \( R \) and an inductor \( L \) in series, the current \( I(t) \) decays exponentially when the circuit is disconnected from a power source or when a switch is opened. The mathematical expression for the decay of current in an RL circuit is derived from the differential equation governing the circuit.
### Derivation
1. **Circuit Description**:
When the switch in an RL circuit is opened or the power source is disconnected, the circuit looks like this:
- **Inductor** \( L \) in series with a **Resistor** \( R \).
2. **Kirchhoff's Voltage Law**:
Applying Kirchhoff's Voltage Law (KVL) around the circuit loop gives:
\[
V_L + V_R = 0
\]
where \( V_L \) is the voltage across the inductor, and \( V_R \) is the voltage across the resistor.
3. **Voltage Across Inductor**:
The voltage across the inductor is given by:
\[
V_L = L \frac{dI(t)}{dt}
\]
4. **Voltage Across Resistor**:
The voltage across the resistor is:
\[
V_R = I(t) R
\]
5. **Combining These**:
Since the sum of voltages in the loop must be zero:
\[
L \frac{dI(t)}{dt} + I(t) R = 0
\]
6. **Solving the Differential Equation**:
Rearranging terms:
\[
L \frac{dI(t)}{dt} = -I(t) R
\]
Dividing through by \( L \):
\[
\frac{dI(t)}{dt} = -\frac{R}{L} I(t)
\]
This is a first-order linear differential equation. Its general solution is:
\[
I(t) = I_0 e^{-\frac{R}{L} t}
\]
where \( I_0 \) is the initial current at \( t = 0 \).
### Summary
The expression for the current \( I(t) \) in an RL circuit decaying over time is:
\[
I(t) = I_0 e^{-\frac{R}{L} t}
\]
where:
- \( I_0 \) is the initial current,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( t \) is time.
This formula indicates that the current decreases exponentially with a time constant \( \tau = \frac{L}{R} \). The time constant \( \tau \) represents the time it takes for the current to decrease to approximately 37% of its initial value.
### Derivation
1. **Circuit Description**:
When the switch in an RL circuit is opened or the power source is disconnected, the circuit looks like this:
- **Inductor** \( L \) in series with a **Resistor** \( R \).
2. **Kirchhoff's Voltage Law**:
Applying Kirchhoff's Voltage Law (KVL) around the circuit loop gives:
\[
V_L + V_R = 0
\]
where \( V_L \) is the voltage across the inductor, and \( V_R \) is the voltage across the resistor.
3. **Voltage Across Inductor**:
The voltage across the inductor is given by:
\[
V_L = L \frac{dI(t)}{dt}
\]
4. **Voltage Across Resistor**:
The voltage across the resistor is:
\[
V_R = I(t) R
\]
5. **Combining These**:
Since the sum of voltages in the loop must be zero:
\[
L \frac{dI(t)}{dt} + I(t) R = 0
\]
6. **Solving the Differential Equation**:
Rearranging terms:
\[
L \frac{dI(t)}{dt} = -I(t) R
\]
Dividing through by \( L \):
\[
\frac{dI(t)}{dt} = -\frac{R}{L} I(t)
\]
This is a first-order linear differential equation. Its general solution is:
\[
I(t) = I_0 e^{-\frac{R}{L} t}
\]
where \( I_0 \) is the initial current at \( t = 0 \).
### Summary
The expression for the current \( I(t) \) in an RL circuit decaying over time is:
\[
I(t) = I_0 e^{-\frac{R}{L} t}
\]
where:
- \( I_0 \) is the initial current,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( t \) is time.
This formula indicates that the current decreases exponentially with a time constant \( \tau = \frac{L}{R} \). The time constant \( \tau \) represents the time it takes for the current to decrease to approximately 37% of its initial value.