Why the need to be careful opening a switch on this RL circuit?

[zenterix]

Homework Statement

Consider the RL circuit shown in the picture below.

Relevant Equations

The switch $S$ is closed for a long time and then at $t=0$ it is opened.

For the period before $t=0$, during which the switch was closed we have

$$-\epsilon_0+IR_1=-L\dot{I}\tag{1}$$

$$\dot{I}+\frac{R_1}{L}I=\frac{\epsilon_0}{L}\tag{2}$$

and the solution to this differential equation is

$$I(t)=\frac{\epsilon_0}{R_1}+e^{-\frac{R_1}{L}t}\tag{3}$$

Thus, $I(\infty)=\frac{\epsilon_0}{R_1}$ which is the current that is flowing at the new $t=0$ when we open the switch.

Let's call this

$$I_0=\frac{\epsilon_0}{R_1}\tag{4}$$

For $t>0$ the differential equation becomes

$$\dot{I}+\frac{R_1+R_2}{L}I=\frac{\epsilon_0}{L}\tag{5}$$

At this point, the problem says to assume that $\epsilon_0$ is approximately $0$. That is, the "battery emf is negligible compared to the total emf around the circuit for times after the switch is opened".

If we do this then we are left with just a homogeneous differential equation and the solution is

$$I(t)=I_0e^{-\frac{(R_1+R_2)t}{L}}\tag{6}$$

The value of the total emf around the circuit is

$$\mathcal{E}_L=-L\dot{I}=I_0(R_1+R_2)e^{-\frac{(R_1+R_2)t}{L}}\tag{7}$$

$$=\epsilon_0\frac{R_1+R_2}{R_1}e^{-\frac{(R_1+R_2)t}{L}}\tag{8}$$

Consider the potential drop across resistor $R_2$.

$$V_+-V_-=I_0R_2=\frac{\epsilon_0 R_2}{R_1}\tag{9}$$

Suppose $R_2=80R_1$..

Then $V_+-V_-=80\epsilon_0$.

My question is about interpreting this last result and also understanding why "you need to be very careful when you open the switch" when $R_2$ is much larger than $R_1$.

At this point I am kinda shocked that the voltage drop on $R_2$ can be 80 times larger than the battery emf.

How can this be?

 

[zenterix]

We can already see it in equation (8).

The back emf at $t=0$ is $\frac{R_1+R_2}{R_1}\epsilon_0$.

As $t$ increases, this back emf goes down to zero.

Initially however, it can be made arbitrarily large by increasing the resistance $R_2$.

I'm looking for an intuitive explanation for what is happening.

 

[gneill]

When the switch is closed for a long time, the current through the inductor will be $I$ ( given by $E_0/R_1$). As soon as the switch is opened, that current through the inductor will persist; the inductor will not change its current rapidly. The inductor will create whatever voltage it needs to maintain that current (initially).

That means the current $I$ will flow through the battery and BOTH resistors. What happens when $R_2$ is much larger than $R_1$? What happens when $R_2$ approaches infinity?

 

[zenterix]

Here are my impressions at this point

1) Right before we open the switch there is a current $I_0$ flowing.

2) When we close it, all of a sudden there is a huge resistance in the circuit. The current wants to decrease immediately but the inductor opposes this change by way of a back emf that is positive.

3) Positive emf means in the direction of positive circulation. This is an emf that induces a current in the same direction as $I_0$.

4) Apparently, at $t=0$ this back emf is $\epsilon_L=81\epsilon_0$.

Since we are assuming that the battery is not in the circuit anymore (it is negligible), most of the voltage drop in the circuit is across resistor $R_2$. In fact, this voltage drop is $80\epsilon_0$.

Now, where does the back emf come from?

From the inductor. Faraday's law tells us that the back emf is proportional to the rate of change of current.

For the same current that was flowing through zero resistance to now flow through resistor $R_2$ a very large potential difference is required.

Apparently, the rate of change of current (which is negative) is large enough (in absolute value) so that at the very instant $t=0$ it sustains the current $I_0$.

Thereafter, the back emf (which is sustaining the current) falls off exponentially and eventually reaches zero at which point there is no current (well, at least in our approximation in which the battery is removed from the analysis; in reality, the current will go down to some positive value determined by the emf provided by the battery).

 

[zenterix]

What happens when R2 is much larger than R1? What happens when R2 approaches infinity?

When $R_2$ is much larger most of the emf is across $R_2$.

When $R_2$ goes to infinity, this emf goes to infinity.

Now, it seems there is a practical danger when the voltage is so high.

 

[gneill]

When $R_2$ is much larger most of the emf is across $R_2$.

When $R_2$ goes to infinity, this emf goes to infinity.

Now, it seems there is a practical danger when the voltage is so high.

Correct!

 

[berkeman]

When $R_2$ goes to infinity, this emf goes to infinity.

Now, it seems there is a practical danger when the voltage is so high.

In a real circuit, you will have parasitic capacitances that will limit the peak back-EMF voltage, but it can still be quite high. The practical problem is that the open circuit voltage across the switch (or relay) will get high enough to cause arcing. Over time, this arcing damages the switch/relay contacts, and can cause the switch/relay to fail with the contacts fused closed.

If the switch is implemented with a transistor, then the $BV_{CEO}$ breakdown voltage can be exceeded by that transient voltage, and the transistor can fail open or closed.

Note that diodes are sometimes used to clamp this transient voltage, to help protect the switch/relay/transistor when the circuit is opened:

https://resources.altium.com/p/using-flyback-diodes-relays-prevents-electrical-noise-your-circuits