What is the formula of current decay in RL circuit?
2 Answers
In an RL circuit (a circuit with a resistor \( R \) and an inductor \( L \)), the current decay after the switch is opened can be described by the formula:
\[
I(t) = I_0 e^{-\frac{R}{L} t}
\]
Where:
- \( I(t) \) is the current at time \( t \),
- \( I_0 \) is the initial current at the moment the switch is opened,
- \( R \) is the resistance in ohms,
- \( L \) is the inductance in henries,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
This equation shows that the current decreases exponentially over time due to the resistance in the circuit. The time constant \( \tau \) for the RL circuit is given by:
\[
\tau = \frac{L}{R}
\]
This time constant indicates how quickly the current decays. After a time period of about \( 5\tau \), the current is generally considered to have decayed to nearly zero.
\[
I(t) = I_0 e^{-\frac{R}{L} t}
\]
Where:
- \( I(t) \) is the current at time \( t \),
- \( I_0 \) is the initial current at the moment the switch is opened,
- \( R \) is the resistance in ohms,
- \( L \) is the inductance in henries,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
This equation shows that the current decreases exponentially over time due to the resistance in the circuit. The time constant \( \tau \) for the RL circuit is given by:
\[
\tau = \frac{L}{R}
\]
This time constant indicates how quickly the current decays. After a time period of about \( 5\tau \), the current is generally considered to have decayed to nearly zero.
In an RL circuit (a circuit containing a resistor and an inductor), the current decay is governed by the natural response of the circuit. When the circuit is de-energized or the switch is opened, the current through the inductor decreases over time. The formula for the current decay in an RL circuit is:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where:
- \( I(t) \) is the current at time \( t \).
- \( I_0 \) is the initial current at \( t = 0 \).
- \( \tau \) (tau) is the time constant of the RL circuit.
The time constant \( \tau \) is given by:
\[ \tau = \frac{L}{R} \]
where:
- \( L \) is the inductance of the inductor in henries (H).
- \( R \) is the resistance of the resistor in ohms (Ω).
### Derivation:
When an RL circuit is disconnected from a voltage source, the current decay follows an exponential curve. The differential equation describing the current \( I(t) \) in the circuit is:
\[ L \frac{dI(t)}{dt} + RI(t) = 0 \]
Solving this differential equation gives:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where the initial condition \( I(0) = I_0 \) is used to determine the constant of integration. The time constant \( \tau = \frac{L}{R} \) represents the time it takes for the current to decay to approximately 36.8% of its initial value.
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where:
- \( I(t) \) is the current at time \( t \).
- \( I_0 \) is the initial current at \( t = 0 \).
- \( \tau \) (tau) is the time constant of the RL circuit.
The time constant \( \tau \) is given by:
\[ \tau = \frac{L}{R} \]
where:
- \( L \) is the inductance of the inductor in henries (H).
- \( R \) is the resistance of the resistor in ohms (Ω).
### Derivation:
When an RL circuit is disconnected from a voltage source, the current decay follows an exponential curve. The differential equation describing the current \( I(t) \) in the circuit is:
\[ L \frac{dI(t)}{dt} + RI(t) = 0 \]
Solving this differential equation gives:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where the initial condition \( I(0) = I_0 \) is used to determine the constant of integration. The time constant \( \tau = \frac{L}{R} \) represents the time it takes for the current to decay to approximately 36.8% of its initial value.