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I am studying a text book and need to come up with time-domain expressions for the current in the resistor \$R\$, the inductor \$L\$ and the potential over the resistor or inductor (which is equivalent).

A switch shorting the current source is opened at t = 0 A switch shorting the current source is opened at t = 0

I have come up with

$$i_R(t) + i_L(t) = I_s = 2A \tag{1}$$

where \$I_s\$ is the source current of \$2A\$.

$$v(t) = L\frac{di_L(t)}{dt} + Ri_R(t) \tag{2}$$

and

$$i_L(t) = K_1e^{\frac{t}{\tau}}, s = \frac{-1}{\tau} \tag{3}$$

which could be written as

$$i_L(t) = K_1e^{st} \tag{4}$$

KVL gives us that the potentials in the resistor-inductor loop sum to zero

$$LsK_1e^{st} = Ri_R(t) \tag{5}$$

and an expression for \$i_R\$ is

$$i_R(t) = 2 - K_1e^{st} \tag{6}$$

I need to solve for \$K_1\$ and \$s\$ to discover \$\tau\$ but I can't seem to solve for either. I substitute 6 into 5 and solve for s to come up with

$$LsK_1e^{st} = R(2 - K_1E^{st}) \tag{7}$$

and

$$s = \frac{R(2 - K_1)}{LK_1} \tag{8}$$

but both \$s\$ and \$K_1\$ remain unknowns.

Conventionally I would solve with initial conditions, at \$t = 0^+\$ I have \$i_R(t) = 2\$ since we know the current in an inductor cannot change instantaneously, and so at the opening of the switch, all the current is delivered to the resistor only.

This gives from equation 6

$$i_R(0^+) = 2 - K_1e^{st} = 2 \tag{9}$$

and so

$$2 = 2 - K_1, K_1 \equiv 0 \tag{10}$$

if I solve for \$t = 0^-\$ I get

$$0 = 2 - K_1, K_1 \equiv 2$$

which is correct but this does not make sense to me as it has derived information from the \$t > 0\$ circuit and solved for the \$t < 0\$ circuit

Where have I gone wrong?

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2 Answers 2

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Conventionally I would solve with initial conditions, at \$t = 0^+\$ I have \$i_R(t) = 2\$ since we know the current in an inductor cannot change instantaneously, and so at the opening of the switch, all the current is delivered to the resistor only.

You should think of the initial condition being the state of the inductor current.

$$i_L(0)=0$$

Since the inductor current doesn't change instantaneously, there's no need to distinguish \$t=0^-\$ from \$t=0^+\$.

which is correct but this does not make sense to me as it has derived information from the \$t > 0\$ circuit and solved for the \$t < 0\$ circuit

You know that \$i_L(0)=0\$ because of information about the \$t<0\$ circuit, not the \$t>0\$ circuit, so there's no contradiction.

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  • \$\begingroup\$ I understand that already, from this I can get \$v(0^+) = LsK_1e^{st} + Ri_R(t)\$ to become \$v(0^+) = Ri_R(t) = 20V\$, I'm lost as to why the \$v(t)\$ is the output potential expression, and is non-zero, but the KVL for the loop gives me that the inductor and resistor potential sum to zero, because the expression for the output potential is the sum of these two, which KVL says is zero... \$\endgroup\$
    – Supernovah
    Commented Aug 9, 2019 at 3:52
  • \$\begingroup\$ @Supernovah, \$i_L(t)\$ is your state variable. Solve for this first. Then \$v(t) = (10\ \Omega)(2\ {\rm A} - i_L(t))\$. \$\endgroup\$
    – The Photon
    Commented Aug 9, 2019 at 4:02
  • \$\begingroup\$ I think I have an answer now, I substituted 6 into 2 after also substituting the potential solution 4 in and initial condition for \$v(0^+) = 20V\$, which gives \$20 = LK_1se^{st} + R(2 - K_1e^{st})\$, simplifying this out I can solve for \$s = 5, \tau \equiv -1/5\$ which I think may be correct. I'll keep working on it... except that my \$\tau\$ is somehow negative... \$\endgroup\$
    – Supernovah
    Commented Aug 9, 2019 at 4:08
  • \$\begingroup\$ Oh I see, my expression for \$i_L\$ is supposed to be for \$i_R\$, I'll keep going... \$\endgroup\$
    – Supernovah
    Commented Aug 9, 2019 at 4:18
  • \$\begingroup\$ Solve for \$i_L\$ first. Then \$i_R(t)=2-i_L(t)\$. \$\endgroup\$
    – The Photon
    Commented Aug 9, 2019 at 4:23
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It turns out there is are multiple errors.

Firstly the model for the inductor current \$i_L(t)\$ is incorrect, and it should be that the resistor current is that which decays to zero.

$$i_R(t)=K_1e^{st}\tag{1}$$

since the sum of the two currents must be the input current (KCL with respect to the net which connects all three elements):

$$i_L(t)=2-i_R(t)=2-K_1e^{st}\tag{2}$$

and the expression for the measured output potential with respect to time would instead be:

$$v(t)=L\frac{di_L(t)}{dt}{\equiv}Ri_R(t)\tag{3}$$

the first differential of the expression for \$i_L(t)\$ gives:

$$\frac{di_L(t)}{dt}=-K_1se^{st}\tag{4}$$

substituting this into the expression for the potential gives:

$$v(t)=-LK_1se^{st}{\equiv}RK_1e^{st}\tag{5}$$

initial conditions for \$t>0\$ indicate that the current in the inductor is zero, and all of the current flows through the resistor, so:

$$-LK_1se^{st}=RK_1e^{st}\tag{6}$$

$$e^{st}{\rightarrow}1,{\quad}-Ls=R,{\quad}s=\frac{-R}{L}=-5,{\quad}\tau=0.2\tag{7}$$

also

$$i_R(0^+)=K_1e^{st},{\quad}e^{st}{\rightarrow}1,{\quad}K_1=2$$

this lets us develop three expressions for each of the currents \$i_R(t)\$, \$i_L(t)\$ and the potential expression \$v(t)\$

$$i_R(t)=2e^{\frac{-t}{0.2}},{\quad}i_L(t)=2-2e^{\frac{-t}{0.2}},{\quad}v(t)=20e^{\frac{-t}{0.2}}$$

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