RL Circuit Analysis- Right After Switch Opens
RL Circuit Analysis When the Switch Opens
Most textbooks show RL circuits with the switch closing. That's the easy part. The interesting stuff happens when the switch opens and the inductor suddenly has nowhere to dump its current.
Here's what actually goes down.
Why Opening a Switch Is Different
When you close a switch, the inductor resists changes in current. It builds up magnetic field gradually. That's textbook behavior.
When you open the switch, the inductor faces a different problem: current was flowing and now it has to stop. An inductor hates sudden current changes. So it generates whatever voltage it needs to keep that current moving—often thousands of volts.
This voltage spike is what destroys contacts, creates arcs, and sends voltage waveforms haywire on your scope.
The Physics in Plain Terms
An inductor stores energy in its magnetic field. The relationship is:
E = ½LI²
When the switch opens:
- The current tries to keep flowing through the same path
- The path suddenly disappears
- The inductor generates a reverse voltage to maintain current
- This voltage can far exceed the source voltage
- Current decays exponentially through whatever path exists
The Time Constant During Discharge
Once the switch opens, you've got an LR circuit discharging. The time constant is still:
τ = L/R
But now R is the total resistance in the discharge path, which might be completely different from the resistance during charging.
Current Decay Equation
i(t) = I₀ × e^(-t/τ)
Where I₀ is the current flowing at the instant the switch opened.
Voltage Across Inductor During Discharge
v(t) = -I₀ × R × e^(-t/τ)
Notice the negative sign. This is because the inductor polarity flips during discharge—it now acts as a temporary voltage source opposing the original direction.
What Actually Happens in the Circuit
Picture this: current I₀ was flowing through the inductor and source. Switch opens. The inductor immediately generates a high voltage to push that same current through any available path—often through the switch contacts, through parasitic capacitance, or through a flyback diode if you're smart enough to include one.
The current doesn't stop instantly. It decays. The voltage doesn't stay constant. It spikes and then decays with the current.
Practical Example
Let's say you have:
- L = 100 mH
- R = 50 Ω
- Source voltage = 12V
- Current before opening = 240 mA
Time constant τ = L/R = 100mH/50Ω = 2 ms
When switch opens, current decays as:
i(t) = 0.24 × e^(-t/0.002)
Voltage across inductor at t=0:
v(0) = -0.24 × 50 = -12V
That doesn't look too bad. But if the only path is through switch contact arcing or parasitic capacitance, R becomes very large, τ becomes tiny, and the initial voltage spike goes through the roof.
Charging vs Discharging: Side-by-Side Comparison
| Parameter | Switch Closes (Charging) | Switch Opens (Discharging) |
|---|---|---|
| Current | Grows from 0 to I₀ | Decays from I₀ to 0 |
| Inductor voltage | Drops from V to 0 | Spikes then decays |
| Energy | Absorbed from source | Released to circuit |
| Time constant | τ = L/R (total) | τ = L/R (discharge path) |
| Initial voltage | V (source voltage) | Can exceed source |
How to Analyze This Step by Step
Step 1: Find Current Before Opening
Use the steady-state condition before the switch opens. Inductor acts like a short, so find current using Ohm's law with the source voltage and series resistance.
Step 2: Identify the Discharge Path
When switch opens, trace where current can flow. This might be:
- Through a flyback diode
- Through contact arcing
- Through parasitic capacitance
- Through a snubber circuit
Step 3: Calculate New Time Constant
Use the resistance in the discharge path. This is often different from the charging resistance.
Step 4: Apply Decay Equations
Current: i(t) = I₀e^(-t/τ)
Inductor voltage: vL(t) = -I₀ × R_discharge × e^(-t/τ)
Step 5: Calculate Peak Voltage
The peak voltage at the instant of opening is:
V_peak = I₀ × R_discharge_path
If R is just the switch contact resistance (very low), voltage can spike to thousands of volts. If R includes a snubber (maybe 1kΩ), voltage might be 240V. Plan accordingly.
Why This Matters in Real Circuits
Inductive kickback when switches open is why:
- Mechanical relay contacts wear out
- Transistors fail without flyback diodes
- Automotive ignition systems work
- Switch-mode power supplies need snubbers
Ignore this analysis and your circuits will fail. It's that simple.
Common Mistakes
Using the wrong R in the time constant. The resistance during discharge is not necessarily the same as during charging. Always identify the actual discharge path.
Forgetting the voltage can exceed source voltage. Inductors don't care about your source voltage during discharge. They generate whatever they need.
Assuming current stops immediately. Current continues to flow at the instant of opening. It decays exponentially, not instantly.
Ignoring the polarity reversal. The inductor voltage sign flips during discharge. This matters for diode orientation and circuit analysis.
The Bottom Line
RL circuit analysis when a switch opens is fundamentally different from closing because the inductor must dispose of stored energy. The current doesn't vanish—it decays through whatever path exists. The voltage doesn't stay at source levels—it spikes based on the discharge resistance.
Calculate the peak voltage. Provide a safe discharge path. Use flyback diodes or snubbers. Otherwise, something in your circuit will fail.