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I tried to look through my textbook to find anything similar to this one and have tried every solution I can think of to no avail. For now, I settled on the fact that I may be over complicating the solution so I calculated the initial and final current of each and used those to find the forced response for t > 0.

The function \$ u(t) \$ is a piecewise function with a value of 0 when \$t < 0\$ and 1 when \$t \geqslant 0 \$. My intial conditions and final conditions are as follows:

$$i_1(0) = i_2(0) = 0 A $$ $$i_1(\infty) = 5A \ \ , \ i_2(\infty) = 2A$$ $$\tau_1 = 0.5\ \ , \ \tau_2 = 0.2$$

These yield the final solutions:

$$i_1(t) = 5(1-e^{-2t})$$ $$i_2(t) = 2(1-e^{-5t})$$

This seems too simple to be true though. I've also tried an approach using mesh analysis but that resulted in constant answers (which seems wrong because it should be a time varying solution?). I also tried using nodal analysis with the node between the 4 and 6 ohm resistors but had no luck.

What approach would be the "correct" one? I don't actually know the correct answer either so any help is greatly appreciated.

P.S. sorry if my math formatting is wrong or if I did not ask my question properly. It is my first time posting on this board and I tried my best to make everything as clear as possible :)

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  • \$\begingroup\$ Just curious, but can you explain how it is that at \$t \to \infty\$ the current in the \$6\:\Omega\$ resistor would be \$2\:\text{A}\$, since \$i_2=2\:\text{A}\$, while also \$i_1=5\:\text{A}\$? Wouldn't that imply that the current in the \$4\:\Omega\$ resistor must be the sum, or \$7\:\text{A}\$ and would therefore drop \$28\:\text{V}\$ across it? \$\endgroup\$
    – jonk
    Commented Mar 1, 2020 at 8:30
  • \$\begingroup\$ I way over simplified my solution in this one. The assumption (which still seems totally wrong to me) was that if you assume that the first inductor remains at its initial conditions then it will essentially operate like an open circuit. One of my solutions that I was re-exploring was to use a mixture of KCL and KVL to solve for a point v(t) and then plugging in initial conditions should yield a di(0)/dt for both of the current. Not sure if I'm on the right track with this since I struggle to convert to an answer. \$\endgroup\$ Commented Mar 1, 2020 at 8:49
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    \$\begingroup\$ I may argue that at \$t\to \infty\$ then \$i_2=0\:\text{A}\$, while \$i_1=5\:\text{A}\$. \$\endgroup\$
    – jonk
    Commented Mar 1, 2020 at 8:52
  • \$\begingroup\$ That definitely makes much more sense than my approach. Do you think this is the correct approach as well? I was imagining that the correct answer would involve some sort of mixture of a forced response for the first and second inductor that might require some differential equation to be solved? \$\endgroup\$ Commented Mar 1, 2020 at 8:59
  • \$\begingroup\$ I don't think there is another consistent answer as \$t\to\infty\$. Do you see another one? \$\endgroup\$
    – jonk
    Commented Mar 1, 2020 at 9:02

3 Answers 3

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I had mentioned that I'd probably take a different approach than user600016's approach, using KCL instead of KVL. Both work for me, but I'm usually more comfortable with KCL as I tend to make fewer errors as I develop the equations. That's a personal detail and may not apply to anyone else. But since it is a different approach than the one taken by user600016, it maybe worth the time to add it here as it provides another way of looking at things. And more ways are better than fewer.

So, let's draw out your schematic as it should have been drawn out in your question, so that we get labeled part numbers. (You should get into the practice of this.)

schematic

simulate this circuit – Schematic created using CircuitLab

I agree, of course, with your initial conditions regarding the currents at \$t=0^-\:\text{s}\$ when \$V_0=0\:\text{V}\$. These currents will be the same at \$t=0^+\:\text{s}\$ when \$V_0=20\:\text{V}\$. We can also say that at \$t=0^+\:\text{s}\$, \$V_1=20\:\text{V}\$ and \$V_2=20\:\text{V}\$, since there are no currents in either resistor and therefore no voltage drops across them.

KCL develops the following two equations:

$$\begin{align*} \frac{V_1}{R_1}+\frac{1}{L_1}\int V_1\:\text{d}t+\frac{V_1}{R_2}&=\frac{V_0}{R_1}+\frac{V_2}{R_2}\\\\ \frac{V_2}{R_2}+\frac{1}{L_2}\int V_2\:\text{d}t&=\frac{V_1}{R_2} \end{align*}$$

(I've placed the out-flowing currents on the left side of the equation and the in-flowing currents on the right side.)

Taking the time derivatives (to remove those integrals):

$$\begin{align*} \left(\frac{1}{R_1}+\frac{1}{R_2}\right)\frac{\text{d}\,V_1}{\text{d}t}+\frac{V_1}{L_1}&=\frac{1}{R_2}\frac{\text{d}\,V_2}{\text{d}t}&&\rightarrow &\frac{\text{d}\,V_1}{\text{d}t}+\frac{R_1\mid \mid R_2}{L_1}V_1&=\frac{R_1}{R_1+R_2}\frac{\text{d}\,V_2}{\text{d}t}\\\\ \frac{1}{R_2}\frac{\text{d}\,V_2}{\text{d}t}+\frac{V_2}{L_2}&=\frac{1}{R_2}\frac{\text{d}\,V_1}{\text{d}t}&&\rightarrow &\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2&=\frac{\text{d}\,V_1}{\text{d}t} \end{align*}$$

Substitution of the second equation into the first (right sides) and then taking the time derivative (again):

$$\begin{align*} \left(\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2\right)+\frac{R_1\mid \mid R_2}{L_1}\int\left(\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2\right)\:\text{d}t&=\frac{R_1}{R_1+R_2}\frac{\text{d}\,V_2}{\text{d}t}\\\\ \left(\frac{\text{d}^2\,V_2}{\text{d}t^2}+\frac{R_2}{L_2}\frac{\text{d}\,V_2}{\text{d}t}\right)+\frac{R_1\mid \mid R_2}{L_1}\left(\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2\right)&=\frac{R_1}{R_1+R_2}\frac{\text{d}^2\,V_2}{\text{d}t^2}\\\\ \frac{\text{d}^2\,V_2}{\text{d}t^2}+\left(\frac{R_1}{L_1}+\frac{R_1+R_2}{L_2}\right)\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_1\, R_2}{L_1\, L_2}V_2&=0\:\frac{\text{V}}{\text{s}^2}\\\\\text{Similarly, you'll wind up with the same expression for }V_1,\\\\\frac{\text{d}^2\,V_1}{\text{d}t^2}+\left(\frac{R_1}{L_1}+\frac{R_1+R_2}{L_2}\right)\frac{\text{d}\,V_1}{\text{d}t}+\frac{R_1\, R_2}{L_1\, L_2}V_1&=0\:\frac{\text{V}}{\text{s}^2} \end{align*}$$

As mentioned before, the initial conditions at \$t=0^+\:\text{s}\$, \$V_1=20\:\text{V}\$ and \$V_2=20\:\text{V}\$. But we also need the initial conditions for \$\frac{\text{d}\,V_1}{\text{d}t}\$ and \$\frac{\text{d}\,V_2}{\text{d}t}\$, which will make the difference for us. Let's assume that \$I_1\$ is the current in \$L_1\$ and that \$I_2\$ is the current in \$L_2\$:

$$\begin{align*} V_1&=V_0-R_1\left(I_1+I_2\right)&&&V_2&=V_1-R_2\,I_2\\\\&&&\therefore\\\\ \frac{\text{d}\,V_1}{\text{d}t}&=\frac{\text{d}\,V_0}{\text{d}t}-R_1\left(\frac{\text{d}\,I_1}{\text{d}t}+\frac{\text{d}\,I_2}{\text{d}t}\right)&&&\frac{\text{d}\,V_2}{\text{d}t}&=\frac{\text{d}\,V_1}{\text{d}t}-R_2\frac{\text{d}\,I_2}{\text{d}t}\\\\&=-R_1\left(\frac{\text{d}\,I_1}{\text{d}t}+\frac{\text{d}\,I_2}{\text{d}t}\right)&&&&=\frac{\text{d}\,V_1}{\text{d}t}-R_2\frac{V_2}{L_2}\\\\&=-R_1\left(\frac{V_1}{L_1}+\frac{V_2}{L_2}\right) \end{align*}$$

We can resolve this at \$t=0^+\:\text{s}\$. Clearly, \$\frac{\text{d}\,V_1}{\text{d}t}=-4\:\Omega\left(\frac{20\:\text{V}}{2\:\text{H}}+\frac{20\:\text{V}}{2\:\text{H}}\right)=-80\:\frac{\text{V}}{\text{s}}\$ and from that we can now solve for \$\frac{\text{d}\,V_2}{\text{d}t}=-80\:\frac{\text{V}}{\text{s}}-6\:\Omega\,\frac{20\:\text{V}}{2\:\text{H}}=-140\:\frac{\text{V}}{\text{s}}\$.

Now you have all four conditions at \$t=0^+\:\text{s}\$: \$V_1\$, \$\frac{\text{d}\,V_1}{\text{d}t}\$, \$V_2\$, and \$\frac{\text{d}\,V_2}{\text{d}t}\$. So you can now easily solve both 2nd order linear diff-eqs for each voltage node and you therefore have \$V_{1\left(t\right)}\$ and \$V_{2\left(t\right)}\$ for \$t\ge 0\:\text{s}\$.

From there, it's a cake walk to get the currents:

$$\begin{align*}I_1&=\frac{V_0-V_1}{R_1}-\frac{V_1-V_2}{R_2}&&&I_2&=\frac{V_1-V_2}{R_2}\end{align*}$$

Post your results and I'll share mine. (Just as an way to check your results, you should find that the peak current in \$L_2\$ will be \$\approx 1.165\:\text{A}\$ and will occur at \$t\approx 358.35\:\text{ms}\$. Those were computed from the formulas using sympy. Below, you can then see the LTspice output, as well. It confirms the equations.)

enter image description here


Appendix

Enough time has passed without a response. So, for others, I'll go ahead and complete the above.

First, let's set up \$\tau_{_1}=\frac{L_1}{R_1}\$, \$\tau_{_2}=\frac{L_2}{R_2}\$, and \$\tau_{_3}=\frac{L_2}{R_1}\$. (I think you can see why these may be the interesting \$\tau\$ values.) Let's also set \$a=\tau_{_1}\,\tau_{_2}\,\tau_{_3}\$ and \$b=\tau_{_1}\,\tau_{_2}+\tau_{_1}\,\tau_{_3}+\tau_{_2}\,\tau_{_3}\$, so that putting these in parallel, as if they were resistances, provides a new \$\tau_{_4}=\frac{\tau_{_1}\,\tau_{_2}\,\tau_{_3}}{\tau_{_1}\,\tau_{_2}+\tau_{_1}\,\tau_{_3}+\tau_{_2}\,\tau_{_3}}=\frac{a}{b}\$. Let's set \$c=\tau_{_3}\$ and set \$\tau_{_5}^2=\sqrt{b^2-4\,a\,c}\$ so that \$\omega_{_0}=\frac{\sqrt{b^2-4\,a\,c}}{2\,a}=\frac{\tau_{_5}^2}{2\,a}\$ (and that \$\tau_{_0}=\frac{1}{\omega_{_0}}\$.)

With the above in hand, the generalized (and specific) answers are (for \$t\ge 0\:\text{s}\$):

$$\begin{align*} V_{1\mathrel{\mkern+3mu\left(t\right)}}&=V_\text{0}\cdot e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\cosh\left(\frac{t}{\tau_{_0}}\right)+\frac{2\,\tau_{_1}\,\tau_{_3}-b}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\\\\&=20\:\text{V}\cdot e^{^{\frac{-7}{2}t}}\left[\cosh\left(\frac{5}{2}t\right)-\frac15\cdot\sinh\left(\frac{5}{2}t\right)\right]\\\\ V_{2\mathrel{\mkern+3mu\left(t\right)}}&=V_\text{0}\cdot e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\cosh\left(\frac{t}{\tau_{_0}}\right)+\frac{-b}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\\\\&=20\:\text{V}\cdot e^{^{\frac{-7}{2}t}}\left[\cosh\left(\frac{5}{2}t\right)-\frac75\cdot\sinh\left(\frac{5}{2}t\right)\right]\\\\ I_{R1\mathrel{\mkern+3mu\left(t\right)}}&=\frac{V_0-V_{1\mathrel{\mkern+3mu\left(t\right)}}}{R_1}=\frac{V_\text{0}}{R_1}\cdot\left\{ 1-e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\cosh\left(\frac{t}{\tau_{_0}}\right)+\frac{2\,\tau_{_1}\,\tau_{_3}-b}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\right\}\\\\&=5\:\text{A}\cdot\left\{ 1- e^{^{\frac{-7}{2}t}}\left[\cosh\left(\frac{5}{2}t\right)-\frac15\cdot\sinh\left(\frac{5}{2}t\right)\right]\right\}\\\\ I_{2\mathrel{\mkern+3mu\left(t\right)}}&=\frac{V_{1\mathrel{\mkern+3mu\left(t\right)}}-V_{2\mathrel{\mkern+3mu\left(t\right)}}}{R_2}=\frac{V_\text{0}}{R_2}\cdot e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\frac{2\,\tau_{_1}\,\tau_{_3}}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\\\\&=4\:\text{A}\cdot e^{^{\frac{-7}{2}t}}\cdot\sinh\left(\frac{5}{2}t\right)\\\\ I_{1\mathrel{\mkern+3mu\left(t\right)}}&=I_{R1\mathrel{\mkern+3mu\left(t\right)}}-I_{2\mathrel{\mkern+3mu\left(t\right)}} \end{align*}$$

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\$20-4(i_1+i_2)-2 \frac{di_1}{dt}=0\$ , \$20-4(i_1+i_2)-6i_2-2\frac{di_2}{dt}=0\$. You just need at \$t=0, i_1=i_2=0\$. (You don't need to explicitly calculate the values at \$t=\infty\$, but still those values you mentioned are incorrect).

Subtracting, we get \$3i_2+\frac{di_2}{dt}-\frac{di_1}{dt}=0\$

Substituting the value of i2 from here to equation 1,

\$10-2i_1 -\frac{5}{3} \frac{di_1}{dt}+\frac{2}{3} \frac{di_2}{dt}=0\$.

Differentiating first equation wrt time, \$2(\frac{di_1}{dt}+\frac{di_2}{dt})+\frac{d^2 i_1}{dt^2}=0\$

So using above two equations, \$\frac{11}{3} \frac{di_1}{dt}-30+6i_1+\frac{d^2 i_1}{dt^2}=0\$.

Now solving this second order differential equation you should have your answer.

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  • \$\begingroup\$ Doesn't Latex work in this site? \$\endgroup\$
    – user600016
    Commented Mar 1, 2020 at 9:55
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    \$\begingroup\$ You need to use '$$' here instead of just '$' like most other sites. \$\endgroup\$
    – PeterJ
    Commented Mar 1, 2020 at 9:58
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    \$\begingroup\$ For in-line you open and close the latex with \$ :) I will try your solution out tomorrow and let you know! \$\endgroup\$ Commented Mar 1, 2020 at 9:59
  • \$\begingroup\$ Just checked this a bit late but I think your equation has an error. It should be \$20-4(i_1+i_2) -2\frac{di_1}{dt} + 2\frac{di_2}{dt} = 0 \$ because both current flow through the center inductor and in opposite directions. This error propagates throughout but I will try to use this method to solve again. \$\endgroup\$ Commented Mar 2, 2020 at 8:48
  • \$\begingroup\$ @Dancing Leaf could you please explain why? I just applied Kirchoff law for the 20V,4 ohm,2H loop. Di2/dt doesn't appear in that. \$\endgroup\$
    – user600016
    Commented Mar 2, 2020 at 8:58
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Well, let's analyze your circuit a bit more:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

When we use and apply KVL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{in}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$


Applying this to your circuit, we need to use (the 'complex' s-domain):

  1. $$\text{R}_2=\text{sL}_1\space\space\space\Longrightarrow\space\space\space\text{R}_2=2\text{s}\tag3$$
  2. $$\text{R}_4=\text{sL}_2\space\space\space\Longrightarrow\space\space\space\text{R}_4=2\text{s}\tag4$$
  3. $$\text{V}_\text{in}=\mathcal{L}_t\left[20\theta\left(t\right)\right]_{\left(\text{s}\right)}=\frac{20}{\text{s}}\tag5$$

I used Mathematica to solve the set of equations, and I used the following code:

FullSimplify[
 Solve[{I1 == I2 + I3, I1 == (Vin - V1)/R1, I2 == (V1)/R2, 
   I3 == (V1 - V2)/R3, I3 == (V2)/R4}, {I1, I2, I3, V1, V2}]]

Using your values, I got:

In[2]:=FullSimplify[
Solve[{I1 == I2 + I3, I1 == ((20/s) - V1)/4, I2 == (V1)/(2*s), 
I3 == (V1 - V2)/6, I3 == (V2)/(2*s)}, {I1, I2, I3, V1, V2}]]

Out[3]={{I1 -> (30 + 20 s)/(6 s + 7 s^2 + s^3), 
I2 -> (10 (3 + s))/(s (6 + 7 s + s^2)), I3 -> 10/(6 + 7 s + s^2), 
V1 -> (20 (3 + s))/(6 + 7 s + s^2), V2 -> (20 s)/(6 + 7 s + s^2)}}

Solving for \$\text{I}_2\$ and \$\text{I}_3\$, gives (after they where transformed back to the time domain):

  1. $$\text{I}_2\left(t\right)=5-\exp\left(-6t\right)-4\exp\left(-t\right)\tag6$$
  2. $$\text{I}_3\left(t\right)=2\exp\left(-6t\right)\left(\exp\left(5t\right)-1\right)\tag7$$
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  • \$\begingroup\$ Jan, could you re-check your process? Compare the results of my I2(t) with your I3(t). Yours shows a maximum peak around 1.375 A. But this is not reflected by LTspice's simulation. The peak my equation shows (taking derivative, solving for zero) shows a maximum peak around 1.165 A. And that is matched by the LTspice simulation. (Similarly, we get different values for the other inductor's current, as well, with similar relationships against LTspice results.) \$\endgroup\$
    – jonk
    Commented Mar 4, 2020 at 19:47
  • \$\begingroup\$ @jonk I checked it, and edited my answer. \$\endgroup\$ Commented Mar 4, 2020 at 19:55
  • \$\begingroup\$ That's much better! The first one matches what I'd written by hand. The second one is in a different form than my hand-written one, but fine. (I hadn't factored out the exp(-6t).) \$\endgroup\$
    – jonk
    Commented Mar 4, 2020 at 19:57

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