Transient Circuit 2 HW: Find i_L & v_L

[STEMucator]

Homework Statement

I felt some more practice for this was needed and I want to make sure I really understand. The circuit is given:

I'm asked to find the current in the inductor just before the switch is closed $( i_L(t = 0^-) )$.

Then I'm asked to find $v_L$ at the instant just after the switch is closed $( v_L(t = 0^+) ).$

Homework Equations

The Attempt at a Solution

So for $t = 0^-$, the switch is open, and so we have a simple series circuit. The circuit is in steady state and the inductor is behaving like a short circuit. Hence the current $i_L(t = 0^-) = 3mA$ because everything is in series. Also, $v_L(t = 0^-) = 0V$.

That seemed simple enough.

Now I need to find $v_L(t = 0^+)$.

So first off, for $t = 0^+$, the inductor remembers it's current from before, so $i_L(t = 0^-) = i_L(t = 0^+) = 3 mA$.

Now since the switch is closed does that mean the current source gets short circuited?

Would that mean $v_R = IR = (3 mA)(2k) = 6V$?

Now KVL would give $v_R - v_L = 0 \Rightarrow v_L(t = 0^+) = 6V$.

Is this even reasonable?

 

[berkeman]

Yes, you are good so far. For t>0, you just have the resistor and inductor in series, with the initial conditions you have shown. The resistor is lossy, so the current decays from 3mA to 0mA over some amount of time (exponentially). Can you write the differential equation that you will use to solve for the time constant and i(t)?

 

[Symmetry777]

Remember the main Idea of an Inductor. An Inductor resists changes in electric current.

A transient event is a short-lived burst of energy (electrons). A transient is an oscillation or a change in electric current. The higher the change of current the more resistance in the inductor coil (microseconds). The coil(s) will spin the electro magnetic flux field in opposition (Impedance).
A surge suppressor, a transient voltage suppressor is mostly made from an inductor. Inductors suppresses voltage spikes or transient voltage surges.

 

[gneill]

One thing to check is the definitions of any voltages on the diagram. $V_L$ is defined as having its + end at the top of the inductor. Make sure that the voltage you calculate via KVL has this definition in mind.

 

[STEMucator]

Yes, you are good so far. For t>0, you just have the resistor and inductor in series, with the initial conditions you have shown. The resistor is lossy, so the current decays from 3mA to 0mA over some amount of time (exponentially). Can you write the differential equation that you will use to solve for the time constant and i(t)?

So what I wrote was actually correct for those two questions? Great! I'm having an off day so I'm glad to hear I did something right.

If I was asked to find the current $i_L(t > 0)$ I would be plugging in to:

$i_L(t > 0) = I_F + (I_I - I_F)e^{\frac{-t}{\tau_L}}$

Where $\tau_L = \frac{R_{th}}{L}$.

I already know $i_L(t = 0^-) = i_L(t = 0^+) = 3mA = I_I$ from before. It can be observed $R_{th} = 2k$ by looking into the terminals of the inductor, and so $\tau_L = \frac{2k}{1H} = 2 \times 10^3 s$.

Now as $t \rightarrow \infty$, $i_L(t \rightarrow \infty) = 0 = I_F$ because of the resistor dissipating the current and the inductor behaves like a short circuit again.

Therefore $i_L(t > 0) = (3mA)e^{\frac{-t}{2 \times 10^3 s}}$

 

[berkeman]

So what I wrote was actually correct for those two questions? Great! I'm having an off day so I'm glad to hear I did something right.

If I was asked to find the current $i_L(t > 0)$ I would be plugging in to:

$i_L(t > 0) = I_F + (I_I - I_F)e^{\frac{-t}{\tau_L}}$

Where $\tau_L = \frac{R_{th}}{L}$.

I already know $i_L(t = 0^-) = i_L(t = 0^+) = 3mA = I_I$ from before. It can be observed $R_{th} = 2k$ by looking into the terminals of the inductor, and so $\tau_L = \frac{2k}{1H} = 2 \times 10^3 s$.

Now as $t \rightarrow \infty$, $i_L(t \rightarrow \infty) = 0 = I_F$ because of the resistor dissipating the current and the inductor behaves like a short circuit again.

Therefore $i_L(t > 0) = (3mA)e^{\frac{-t}{2 \times 10^3 s}}$

Re-check your time constant there scooter! :-)

http://en.wikipedia.org/wiki/RL_circuit

.

 

[STEMucator]

Re-check your time constant there scooter! :)

http://en.wikipedia.org/wiki/RL_circuit

.

Ahaha I can't even believe that happened. Feels too movie like to be true.

$\tau_L = 0.5 ms$

One thing to check is the definitions of any voltages on the diagram. $V_L$ is defined as having its + end at the top of the inductor. Make sure that the voltage you calculate via KVL has this definition in mind.

I always go clockwise. Is my voltage wrong or something?

$v_L + v_R = 0 \Rightarrow v_L(t = 0^+) = -6V$

 

[gneill]

I always go clockwise. Is my voltage wrong or something?

$v_L + v_R = 0 \Rightarrow v_L(t = 0^+) = -6V$

That's better :)

 

[STEMucator]

That's better :)

Awesome. I think I get it all conceptually now. Should be better after a sleep.

Tomorrows just another day right.