# Find the inductance of an inductor in a circuit

• Jaccobtw

#### Jaccobtw

Homework Statement
An inductor, two resistors, a make-before-break switch, and a battery are connected as shown. The switch throw has been at contact e for a long time and the current in the inductor is 2.5A. Then, at t=0, the switch is quickly moved to contact f. During the next 45ms the current in the inductor drops to 1.5A. If the resistors are both have value 0.4Ω, what is the value of the inductance L in H?
Relevant Equations
Kirchhoff's loop rule
$$I(t) = \frac{\varepsilon}{R}(1 - e^{\frac{-Rt}{L}})$$ When the current has been running for a while we can treat the circuit connected to e like a battery with two resistors. The current at this point is 2.5 A so using kirchhoff's loop rule we find that the emf of the battery is 2V. $$\varepsilon - IR - IR = 0$$
When the switch flips to f, we can use the current function to find L since we know the current (1.5 A) at a particular time. $$I(t) = \frac{\varepsilon}{R}(1 - e^{\frac{-Rt}{L}})$$
Plug in 1.5 for I, 2 for ##\varepsilon##, 0.4 for R, 45 ms for t and solve for L. Where did I go wrong? Thank you

Why do you think the initial current is ##\varepsilon/R##?

• Jaccobtw and Delta2
Why do you think the initial current is ##\varepsilon/R##?
Oops. that should be ##\varepsilon/2R## I think.

• Delta2
Why? Please motivate. Otherwise we don’t know if you have understood or if you are guessing wildly.

• Jaccobtw
Yes that's the twist of this problem, the initial current is ##\frac{\mathcal{E}}{2R}## but the time constant is ##\frac{L}{R}##.

I remember a similar problem was in my exams for entrance to the university 30 years ago and I did the same mistake. Consequence : I didn't pass in the EE department but in the Math departm (it wasn't only that unfortunately I failed even worst in the chemistry exams). Needed very high grades to pass to the EE university at that era.

• Jaccobtw
Yes that's the twist of this problem, the initial current is E2R
The important realisation being that this is given to us by the problem:
Homework Statement:: ... and the current in the inductor is 2.5A. Then, at t=0, ...
There is no need to ever compute ##\mathcal E_0## in this problem. Its introduction and the statement that both resistors have the same resistance are red herrings.

The moment the switch closes the d-f connection the left and right loops become independent and the initial current through the inductor is 2.5 A. That the emf is disconnected when the d-e connection is lost is therefore irrelevant, making the entire left loop irrelevant to the actual analysis.

• Jaccobtw
Plug in 1.5 for I, 2 for ε, 0.4 for R, 45 ms for t and solve for L. Where did I go wrong? Thank you

Have you tried checking that the equation you're using are suitable for computing the answer you're asking for?$$\text{Either of the two}~~~~~~~~~~~I(t) = \frac{\varepsilon}{R}(1 - e^{\frac{-Rt}{L}})~~~~~~\text{or}~~~~~I(t) = \frac{\varepsilon}{2R}(1 - e^{\frac{-Rt}{L}})$$

• Jaccobtw and Delta2
Yes there is no need to compute ##\mathcal{E}##, just to be able to realize what I call here the silent and hidden assumption (or i better say fact than assumption): That the current remains the same before and after the movement of switch, it just changes path through the circuit but remains the same in value, it doesn't mysteriously rises to ##\frac{\mathcal{E}}{R}##.

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• Jaccobtw
Have you tried checking that the equation you're using are suitable for computing the answer you're asking for?$$\text{Either of the two}~~~~~~~~~~~I(t) = \frac{\varepsilon}{R}(1 - e^{\frac{-Rt}{L}})~~~~~~\text{or}~~~~~I(t) = \frac{\varepsilon}{2R}(1 - e^{\frac{-Rt}{L}})$$
I guess its not suitable because I still got the wrong answer. I'm not sure why it'd be unsuitable. I plugged in 1.5 for I, 2.5 for ##\varepsilon/R##, 0.4 for the exponential R, and 45 X 10^-3 s for t. I got about 0.02 for L.

I also get 0.02 approximately (0.0196444 to be more accurate).

Aaaargh we both did another mistake. Check again the equation ##I(t)## is has another mistake (hint: it does not satisfy the condition ##I(0)=2.5##

• Jaccobtw
I’d like to chip-in if I may.

The formula ##I(t) = \frac{\mathscr E}{R}(1 – e^{\frac{-Rt}{L}})## is for an inductor ‘charging’ (in the sense of storing-up energy in its magnetic field). ##R## is the total charging-circuit resistance. During this process the current through the inductor increases reaching a maximum (asymptotic) value of ##\frac{\mathscr E}{R}##.

But the problem requires you to consider the inductor ‘discharging’! During this process the current through the inductor decreases. The initial current (##I_0=2.5A##) must decay to zero.

So (apart from the ‘R’ and ‘2R’ confusion) you are using the wrong formula!

Also, you might find it useful to note that rather then use:
\varepsilon which gives ##\varepsilon##
you can use:
\mathscr E which gives ##\mathscr E##.
to get the usual emf symbol (which my school physics teacher used to call 'curly E').

• • Orodruin, Jaccobtw and Delta2

• Jaccobtw
Why? Please motivate. Otherwise we don’t know if you have understood or if you are guessing wildly.
The initial current should be ##\varepsilon/2R## because there were two resistors in series in the original circuit.

I’d like to chip-in if I may.

The formula ##I(t) = \frac{\mathscr E}{R}(1 – e^{\frac{-Rt}{L}})## is for an inductor ‘charging’ (in the sense of storing-up energy in its magnetic field). ##R## is the total charging-circuit resistance. During this process the current through the inductor increases reaching a maximum (asymptotic) value of ##\frac{\mathscr E}{R}##.

But the problem requires you to consider the inductor ‘discharging’! During this process the current through the inductor decreases. The initial current (##I_0=2.5A##) must decay to zero.

So (apart from the ‘R’ and ‘2R’ confusion) you are using the wrong formula!

Also, you might find it useful to note that rather then use:
\varepsilon which gives ##\varepsilon##
you can use:
\mathscr E which gives ##\mathscr E##.
to get the usual emf symbol (which my school physics teacher used to call 'curly E').
Ah yes! should have used $$I(t) = \frac{\varepsilon}{R} e^{\frac{-Rt}{L}}$$

• alan123hk and Steve4Physics
Ah yes! should have used $$I(t) = \frac{\varepsilon}{R} e^{\frac{-Rt}{L}}$$
actually that ##R## below ##\epsilon## should be ##2R## you can use:
\mathscr E which gives E.
Talking of TeXnicalities, I think \mathcal E ##\mathcal E## is more like what is used in the OP’s image.

Ah yes! should have used $$I(t) = \frac{\varepsilon}{R} e^{\frac{-Rt}{L}}$$
Or just not computed ##\mathcal E## at all as the constant in front (which is ##\mathcal E/2R##) necessarily is 2.5 A from the problem statement itself.

• Steve4Physics
Talking of TeXnicalities, I think \mathcal E ##\mathcal E## is more like what is used in the OP’s image.
Interesting. I've not come across \mathcal. Will check it out.

Or just not computed ##\mathcal E## at all as the constant in front (which is ##\mathcal E/2R##) necessarily is 2.5 A from the problem statement itself.
Yes - the relevant formula is simply ##I(t) = I_0e^{\frac{-Rt}{L}}## with ##I_0##=2.5A.

Talking of TeXnicalities, I think \mathcal E ##\mathcal E## is more like what is used in the OP’s image.

Or just not computed ##\mathcal E## at all as the constant in front (which is ##\mathcal E/2R##) necessarily is 2.5 A from the problem statement itself.
actually that ##R## below ##\epsilon## should be ##2R## right. I just copied the function from the text. I think R originally means the total resistance in the circuit instead of resistance from a single resistor, but obviously the way the problem is set up makes out R = 0.4 ##\Omega##.

If we write the differential equation for the right loop containing one resistor R and the inductor L (after the switch has go to the position f), it will be (using KVL) $$IR+L\frac{dI}{dt}=0$$
Treating this equation purely mathematically, we can find that the general solution is $$I(t)=Ce^{-\frac{R}{L}t}$$ where C can be any constant.
In order to fully solve it we need an initial condition on I, that is a condition $$I(t_0)=I_0$$ so that we can determine the constant C by solving the equation $$I(t_0)=I_0\Rightarrow Ce^{-\frac{R}{L}t_0}=I_0\Rightarrow C=I_0e^{\frac{R}{L}t_0}$$.

Sometimes the initial condition comes automatically (like I(0)=0 for example) but some other times we have to think a bit about the physics of the problem and this is the case here. We have to think a bit in order to set $$I(0)=\frac{\mathcal{E}}{2R}=2.5A$$.