23.10 RL Circuits

RL Circuits

An RL circuit contains a resistor (R) and an inductor (L) connected to a voltage source. The core idea is that current in an RL circuit doesn't change instantly. Instead, it rises or falls along a smooth exponential curve because the inductor resists changes in current. Understanding this time-dependent behavior is essential for analyzing how real circuits respond when you flip a switch.

Current Behavior in RL Circuits

When you connect an RL circuit to a voltage source, the inductor generates a back emf that opposes the rising current. This means the current starts at zero and gradually climbs toward its maximum value:

$I_{max} = V/R$

Think of it like pushing a heavy cart: it doesn't go from stopped to full speed instantly. The inductor's magnetic field needs time to build up.

When you disconnect the voltage source, the reverse happens. The inductor's collapsing magnetic field tries to keep current flowing, so the current decreases exponentially from $I_{max}$ down to zero.

Both of these behaviors trace exponential curves, and the rate at which they happen depends on the circuit's time constant.

Time Constant of RL Circuits

The time constant $\tau$ (tau) tells you how fast the circuit responds to changes. It's calculated as:

$\tau = L/R$

• $L$ = inductance in henries (H)
• $R$ = total resistance in ohms ($\Omega$)

What $\tau$ physically represents:

• When connecting to a voltage source, $\tau$ is the time for the current to reach about 63.2% of $I_{max}$
• When disconnecting, $\tau$ is the time for the current to drop to about 36.8% of its initial value

A larger $L$ or smaller $R$ gives a bigger $\tau$, meaning a slower response. A smaller $L$ or larger $R$ gives a smaller $\tau$ and a faster response. After about $5\tau$, the circuit is effectively at its final state (the current has reached ~99% of its target value).

Current Calculations at Specific Times

You can find the current at any time $t$ using these two equations:

Connecting (current rising):

$I(t) = I_{max}(1 - e^{-t/\tau})$

Disconnecting (current falling):

$I(t) = I_{max} \cdot e^{-t/\tau}$

Where:

• $I_{max} = V/R$ (the steady-state current)
• $e \approx 2.718$ (Euler's number)
• $t$ = time elapsed since the switch was flipped
• $\tau = L/R$

Example walkthrough: Find the current 5 ms after connecting a 12 V battery to an RL circuit with $L = 100 \text{ mH}$ and $R = 50 \text{ } \Omega$.

1. Calculate $I_{max}$: $I_{max} = V/R = 12/50 = 0.24 \text{ A}$

2. Calculate $\tau$: $\tau = L/R = 0.100/50 = 0.002 \text{ s} = 2 \text{ ms}$

3. Plug into the rising-current equation: $I(0.005) = 0.24(1 - e^{-0.005/0.002})$

4. Simplify the exponent: $-0.005/0.002 = -2.5$

5. Evaluate: $I = 0.24(1 - e^{-2.5}) = 0.24(1 - 0.0821) = 0.24 \times 0.918 = 0.220 \text{ A}$

Since 5 ms is 2.5 time constants, the current has already reached about 92% of its maximum.

Electromagnetic Induction in RL Circuits

The transient behavior of RL circuits is a direct consequence of Faraday's law. As the current changes, the magnetic flux through the inductor changes with it. That changing flux induces a back emf in the inductor, which opposes the change in current. This is why the current can't jump to its final value instantly.

The transient response refers to the circuit's behavior during the transition between its initial state and its final (steady) state. During this period, the inductor actively stores or releases energy in its magnetic field. Once enough time has passed (roughly $5\tau$), the current stabilizes. In a DC circuit at steady state, the inductor acts like a plain wire since $dI/dt = 0$ and the voltage across it drops to zero.