What is the time constant of an LR circuit?
2 Answers
The time constant of an LR circuit, which consists of an inductor \( L \) and a resistor \( R \) in series, is a measure of how quickly the current in the circuit builds up or decays when a voltage is applied or removed.
### Definition of Time Constant
In an LR circuit, the time constant \( \tau \) is given by:
\[ \tau = \frac{L}{R} \]
where:
- \( L \) is the inductance of the inductor (in henrys, H)
- \( R \) is the resistance of the resistor (in ohms, Ω)
### Explanation
- **Charging Phase (Current Build-Up)**: When a voltage source is suddenly applied to the LR circuit, the current does not instantly reach its maximum value. Instead, it grows exponentially according to the time constant. The current \( I(t) \) at time \( t \) after the voltage is applied is given by:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
Here, \( V \) is the applied voltage, and \( e \) is the base of the natural logarithm.
- **Discharging Phase (Current Decay)**: If the voltage source is suddenly removed, the current decays exponentially with time. The current \( I(t) \) at time \( t \) after the voltage is removed is given by:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where \( I_0 \) is the current at the moment the voltage was removed.
### Physical Interpretation
- **Short Time Scales**: On time scales much shorter than \( \tau \), the inductor resists changes in current, and the circuit behaves almost like an open circuit.
- **Long Time Scales**: On time scales much longer than \( \tau \), the inductor behaves like a short circuit, and the current reaches its steady-state value or decays completely, depending on whether the voltage is applied or removed.
### Example
If you have an inductor with an inductance of 2 H and a resistor with a resistance of 4 Ω, the time constant \( \tau \) would be:
\[ \tau = \frac{L}{R} = \frac{2 \text{ H}}{4 \text{ Ω}} = 0.5 \text{ s} \]
So, the current in the circuit will reach approximately 63.2% of its final value in 0.5 seconds after the voltage is applied, and it will decay to approximately 36.8% of its initial value in the same time after the voltage is removed.
### Definition of Time Constant
In an LR circuit, the time constant \( \tau \) is given by:
\[ \tau = \frac{L}{R} \]
where:
- \( L \) is the inductance of the inductor (in henrys, H)
- \( R \) is the resistance of the resistor (in ohms, Ω)
### Explanation
- **Charging Phase (Current Build-Up)**: When a voltage source is suddenly applied to the LR circuit, the current does not instantly reach its maximum value. Instead, it grows exponentially according to the time constant. The current \( I(t) \) at time \( t \) after the voltage is applied is given by:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
Here, \( V \) is the applied voltage, and \( e \) is the base of the natural logarithm.
- **Discharging Phase (Current Decay)**: If the voltage source is suddenly removed, the current decays exponentially with time. The current \( I(t) \) at time \( t \) after the voltage is removed is given by:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where \( I_0 \) is the current at the moment the voltage was removed.
### Physical Interpretation
- **Short Time Scales**: On time scales much shorter than \( \tau \), the inductor resists changes in current, and the circuit behaves almost like an open circuit.
- **Long Time Scales**: On time scales much longer than \( \tau \), the inductor behaves like a short circuit, and the current reaches its steady-state value or decays completely, depending on whether the voltage is applied or removed.
### Example
If you have an inductor with an inductance of 2 H and a resistor with a resistance of 4 Ω, the time constant \( \tau \) would be:
\[ \tau = \frac{L}{R} = \frac{2 \text{ H}}{4 \text{ Ω}} = 0.5 \text{ s} \]
So, the current in the circuit will reach approximately 63.2% of its final value in 0.5 seconds after the voltage is applied, and it will decay to approximately 36.8% of its initial value in the same time after the voltage is removed.