- #1

zenterix

- 555

- 76

- 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?