I have the following circuit (see picture 1) where the switch \$ 0 - 1 - 2 \$ is in the "state" \$ 0 - 2\$. The values are \$ R_1 = 20~\Omega \$ , \$ R_2 = 30~\Omega \$, \$ R_3 = 10~\Omega \$, \$ I_q = 4~\mathrm{A} \$, and \$ L = 80~\mathrm{mH} \$.

Picture 1

I have to find the equivalent voltage source to determine the first order ODE for the inductor current \$ i_L(t) \$. This was the note from the professor.

So I tried to do this like in picture 2.

Picture 2

I replaced the current source \$ I_q \$ and the parallel resistor \$ R_2 \$ (both in picture 1) with the voltage source \$ U_q \$ and the series resistor \$ R_2 \$. I know that \$ U_q = I_q \cdot R_2 \$.

But there is an example in the script of the professor where the inductor \$ L \$ is removed and replaced with two clamps (see picture 3).

Picture 3

As an intermediate step to determine the ODE I have to calculate the voltage between the clamps \$ U_{kl} \$ and the internal resistance \$ R_i \$.

Picture 4

I tried to calculate the internal resistance \$ R_i = R_2 + R_3 = 30~\Omega + 10~\Omega = 40~\Omega \$ (see picture 4). This seems to be accurate because the solution, which was given from the professor is \$ 40~\Omega \$. But I'm not sure how to calculate \$ U_{kl} \$ and if I even did the transformation to the equivalent voltage source correctly, because the solution for \$ U_{kl} \$ is \$ 120~\mathrm{V} \$ but I got this for \$ U_q \$.

Can it be that \$ U_q = U_{kl} \$ ?


2 Answers 2


Everything you did was correct, except the last step. Starting from the last drawn circuit we get (\$i = i_{L}\$): $$ U_2 = R_i \cdot i + u_{kl}$$ where \$U_2 = R_2\cdot I_2\$ and \$R_i = R_2 + R_3\$. \$i = i(t)\$ and \$u_{kl} = u_{kl}(t)\$ are the unknown current and voltage we want to determine. Notice that we can express \$u_{kl}\$ in terms of \$i\$ via the equation for an ideal inductor $$ u_{kl}(t) = L\cdot \frac{di}{dt}.$$ Substitution of \$u_{kl}\$ in our initial voltage loop equation leads to $$U_2 = R_i \cdot i + L\cdot \frac{di}{dt},$$ the desired ODE describing the evolution of \$i(t)\$. If we now want to calculate the current \$i\$, we need an additional piece of information. Namely the initial current through the inductor at the beginning of our "observation period". Let \$i_0\$ the initial current through the inductor. In that case the solution to the initial value problem is given by $$ i(t) = \frac{U_2}{R_i} + (i_0 - \frac{U_2}{R_i})\cdot e^{-\frac{(t-t_0)}{\tau}}$$ with time constant \$\tau = \frac{L}{R_i}\$ and starting time of our "observation" \$t_0\$.

The usual approach to deriving the ODEs for circuits, like the one given by you, starts by writing down all the relevant independent voltage loop and node equations. Next, express all currents and voltages in terms of

1.) The inductor currents and voltages

2.) The capacitor currents and voltages

3.) The source currents and voltages

Finally, substitute $$ u_{ind} = L\cdot\frac{di_{ind}}{dt}$$ and $$i_{cap} = C\cdot\frac{du_{cap}}{dt}$$ and you will get the desired system of ODEs, which describes the entire system. It is the same approach as taken here. Further manipulation may be neccessary if you want to extract the ODE for a single quantity from the system of ODEs produced by above approach.

  • \$\begingroup\$ Thank You very much for the reply! I still don't understand two things: 1. You said following from the last circuit we get \$ U_2 = R_i \cdot i + u_{kl} \$, but in the last circuit I have only \$ U_q \$, \$ R_i \$ and \$ u_{kl} \$ 2. The professors answer for \$ U_{kl} \$ is \$ 120 \ V \$ but here in Your answer this is the unknown voltage. For the ODE the professor gives the following answer: \$ \frac{di_L(t)}{dt} + \frac{R_i}{L} \cdot i_L(t) = \frac{U_{kl}}{L} \$. Comparing this to Your answer it seems like what the professor "names" \$ U_{kl} \$ You "name" it \$ U_2 \$. \$\endgroup\$
    – syphracos
    May 17 at 9:15
  • \$\begingroup\$ Could You maybe clarify it because it confuses me a little bit. It seems like I still don't understand what exactly is happening \$\endgroup\$
    – syphracos
    May 17 at 9:17
  • \$\begingroup\$ After some looking and analyzing it, can it be that by \$ u_{kl} \$ You mean the inductor voltage \$ u_L \$ ? \$\endgroup\$
    – syphracos
    May 17 at 9:49

Regarding the answer from @Yggdrasil :

I think I understand it now.

The circuit in picture 2 is actually

picture 5

see the currents on \$ R_2 \$ and \$ R_3 \$. Because the clamps \$ kl\$ are in this case like an open circuit, the currents \$ i_{R_2} = i_{R_3} = 0 \ A \$. It follows then that the CLAMP voltage \$ u_{kl}(t) = u_q(t) = 120 \ V\$.

Then we have the circuit in picture 6:

picture 6

The voltage \$ u_q \$ from picture 3 is now corrected to the clamp voltage \$ u_{kl} \$. The internal resistance is \$ R_i = R_2 + R_3 = 30 \ \Omega + 10 \ \Omega = 40 \ \Omega \$.

And now we have the simplified circuit including the inductor \$ L \$ in picture 7:

picture 7

Here we have \$ u_{kl}(t) = u_{R_i} + u_L = R_i \cdot i(t) + u_L (t) \$, just like in @Yggdrasil's answer.

Then we proceed to get the ODE like @Yggdrasil wrote it.


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