# Finding some variables in a second-order circuit?

Can any one please help me solving these questions .. I always get a wrong answers :/ The first order circuit is much easier than this one !!

for the first time i thought it will not be easy .. but after understanding the concept .. it become much easier ; how ever .. I solved all of them but i need one to check the answers .. Specially (c) :

a) 12/7 * 5 = 60/7 =v(0-)=v(0+)

b) 12/7 =i(0-)=i(0+)

Now for (c) i added the inductors in the formula .. but i'm not sure if it is right :

(c) i(t)=12/7 e^(-t/(3*(0.1m+0.3))) !

and for the final

(d) Vc(inf.)=12 V !

• What have you been able to figure out yourself so far? How do you get the wrong answers that you're getting? – Nick Alexeev Aug 19 '16 at 19:58
• @NickAlexeev My main problem is with 0- and 0+ , I can't see the different between them when I answer the questions ! – Hashim Aug 19 '16 at 20:05
• When solving for 0-, remove the switch from the circuit. When solving for 0+, treat the switch as a wire. What do you get in the two cases? We are not going to answer your questions for you, but if you provide us your full answers, we can point out errors. – Justin Aug 19 '16 at 20:13
• Try looking at this: electronics.stackexchange.com/questions/216677/… If you do some searching there are also plenty of questions and answers to simmilar problems if you search for switch initial condition – Voltage Spike Aug 19 '16 at 20:19
• The problem states that "the switch is closed for a long time before it is opened at time t = 0". So solve for t = 0- with the switch closed and the circuit at DC equilibrium. Then solve for t = 0+ with switch opened using the knowledge that voltage across a capacitor and the current through an inductor are continuous (same as t=0-). With the switch open, the circuit on the right with L2 breaks off to a 1st-order system. It would have been a lot more complicated to solve for (c), the transient of L2, with the switch going from open to close (looks like it would be more than 2nd-order). – rioraxe Aug 19 '16 at 20:31

Let's start with (d), since that's what I brought up and it's easy. The switch is opened at $t=0$ and then you are asked what the voltage across $C_1$ is at $t=+\infty$. At this time, there is no more current into or out of $C_1$, the drop across $R_1$ will be $0V$ and $L_1$ will also have $0V$ across it because $\frac{dI}{dt} = 0$ after all that time. So $V_1$ must equal the voltage across $C_1$, so you are right, as $V_c\left(t=+\infty\right) = 12V$.

Back to (a). At this point, the switch is engaged for a very long time, as well. So again, $\frac{dI}{dt} = 0$ and $\frac{dV}{dt} = 0$ everywhere. This means no voltage drops across inductors, so they are all effectively dead shorts and you can replace them with wires; and no current through capacitors, so they are all effectively open circuits and you can mentally remove them. $L_2$ and $L_3$ now bypass $R_3$. This leaves just $R_1$ in series with $R_2$, across $V_1$, forming a voltage divider. The capacitor voltage will then be $V_c\left(t=0^-\right) = 12V\cdot\frac{5\Omega}{5\Omega+2\Omega} = \frac{60}{7}\Omega = 8.57142857\Omega$.

Now to (b). The inductance current doesn't undergo instantaneous change. So $I_L\left(t=0^+\right) = I_L\left(t=0^-\right) + dI_L$. But $dI_L = \frac{dV\cdot dt}{L}$ and a pair of multiplied infinitesimal values isn't just infinitesimally small, in this context. It's $0A$. So there is $I_L\left(t=0^+\right) = I_L\left(t=0^-\right) = 0A$.

Finally to (c). The current through $L_2$ at $t=0^-$ is indeed $\frac{12V}{2\Omega+5\Omega} = \frac{12}{7}A$. The voltage across $L_2$ is $0V$, so no current is flowing through $R_3$ and also so no current is flowing through $L_3$. So $I_{L_2}\left(t=0^-\right) = \frac{12}{7}A$ and $I_{L_3}\left(t=0^-\right) = 0A$ and $I_{R_3}\left(t=0^-\right) = 0A$.

However, when the switch is opened at $t=0$, then the current through $L_2$ must instantly now continue through the only remaining path, which is through $L_3$ and then $R_3$. But this is impossible, as it would require an infinite voltage to attempt such an infinitely fast change in current in $L_3$. The equation would be something like, $\vert dV\vert = L_3\cdot\vert\frac{0A - \frac{12}{7}A}{dt}\vert$, but given that $dt$ is infinitely small when moving from $t=0^-$ to $t=0^+$, the result is that $dV$ is infinitely large.

It's not possible to write an accurate equation for $I_{L_2}\left(t\right)$.

Note: One caveat in all of this that I need to add. I'm a hobbyist and I have never had so much as a single course of training in electronics or electricity. So I will defer to any professional disagreeing with me about the above.