DE of an LR circuit

Question

Suppose di/dt + 20i = 5 is a DE that models an LR circuit, with i(t) representing the current at a time t in amperes, and t representing the time in seconds. If the resistance of the circuit is 60 Ohms, then what is the inductance in Henries?

I am pretty familiar with finding the general solution and/or setting up IVP's for circuits, but I'm having trouble working backwards for this one. So far I have VL + VR = E(t), and VL = L(di/dt), VR = i(t)R. I have been stumped for an hour on this question and would appreciate any sort of help. Thanks!

Answer 1

Compare your differential equation $$\frac{di}{dt}+20i=5$$ with the general differential equation of an LR-circuit $$\frac{di}{dt}+\frac{R}{L}i=\frac{V}{L}$$ Now what tells $\frac{R}{L}=20$ and $\frac{V}{L}=5$ you?

Answer 2

Assuming you have a constant voltage $V$ driving a series connection of a resistor $R$ and inductor $L$ the differential equation for $i$ will be ${d i (t) \over dt} + {R \over L} i(t) = {V \over L}$.

You are given $R$ hence you can compute the value of $L$ from the coefficient of $i(t)$.

Answer 3

Well, we can set up a system of known equations (assuming a series RL-circuit with a DC source):

$$ \begin{cases} \text{V}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{R}}\left(t\right)+\text{V}_{\space\text{L}}\left(t\right)\\ \\ \text{V}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)\cdot\text{R}\\ \\ \text{V}_{\space\text{L}}\left(t\right)=\text{I}_{\space\text{L}}'\left(t\right)\cdot\text{L}\\ \\ \text{I}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{L}}\left(t\right) \end{cases}\tag1 $$

Which gives:

$$\text{V}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$ $$\text{I}_{\space\text{in}}\left(t\right)=\frac{\text{V}_{\space\text{in}}\left(t\right)}{\text{R}}\cdot\left\{1-\frac{1}{\exp\left(\frac{\text{R}}{\text{L}}\cdot t\right)}\right\}+\frac{\text{I}_{\space\text{in}}\left(0\right)}{\exp\left(\frac{\text{R}}{\text{L}}\cdot t\right)}\tag2$$

And your DE gives:

$$\text{I}_{\space\text{in}}'\left(t\right)+20\cdot\text{I}_{\space\text{in}}\left(t\right)=5\space\Longleftrightarrow\space\text{I}_{\space\text{in}}\left(t\right)=\frac{1}{4}\cdot\left\{1-\frac{1}{\exp\left(20\cdot t\right)}\right\}+\frac{\text{I}_{\space\text{in}}\left(0\right)}{\exp\left(20\cdot t\right)}\tag3$$

So, when $\text{R}=60$, we need to solve:

$$\frac{\text{V}_{\space\text{in}}\left(t\right)}{60}\cdot\left\{1-\frac{1}{\exp\left(\frac{60}{\text{L}}\cdot t\right)}\right\}+\frac{\text{I}_{\space\text{in}}\left(0\right)}{\exp\left(\frac{60}{\text{L}}\cdot t\right)}=\frac{1}{4}\cdot\left\{1-\frac{1}{\exp\left(20\cdot t\right)}\right\}+\frac{\text{I}_{\space\text{in}}\left(0\right)}{\exp\left(20\cdot t\right)}\tag4$$

So, we can see:

$$ \begin{cases} \frac{\text{V}_{\space\text{in}}\left(t\right)}{60}=\frac{1}{4}\\ \\ \frac{60}{\text{L}}\cdot t=20\cdot t \end{cases}\space\space\space\space\space\space\space\space\space\space\space\space\therefore\space\space\space\space\space\space\space\space\space\space\space\space\begin{cases} \text{V}_{\space\text{in}}\left(t\right)=15\\ \\ \text{L}=3 \end{cases}\tag5 $$