Inductor voltages and currents

[PhysicsTest]

Homework Statement

a. What is the energizing circuit time constant?
b. Close the switch and determine the equation for iL and vL during current buildup.
c. What are iL and vL at t = 25 ms?

Relevant Equations

i = E/R*(1 - e^{-Rt/L})

I have redrawn the circuit as below

40 - (I2+0.5)200 - (I2 + IL)*300 = 0 -> eq1
-(I2 + IL)*300 - IL*280 - 4*dIL/dt = 0 -> eq2
Are my equations correct?

 

[DaveE]

Looks good to me.

 

[PhysicsTest]

I tried the following but answer does not match

$40 - (I_2 + 0.5)200 - (I_2+I_l)*300=0$ -> 1
$500i_2 = 140 - 300i_l$ -> 1i
$-(I_2+I_l)*300 - I_l*280 - 4\frac{dI_l} {dt} =0$ -> 2
$-300I_2 - 580I_l - 4 \frac{dI_l} {dt} = 0$ -> 2i
Substituting 2i in 1i
$\frac{di_l} {dt} + 75i_l + 21 =0$ Solving for il
$i_l = \frac{21e^{-75t}} {75} - \frac{21} {75}$
a. Energizing circuit time out, i tried to calculate $\tau$
$\tau = \frac{1} {75} = 0.013$ But the answer is 10ms.

 

[DaveE]

There's a sign error when you put the numbers into eqn 1.
40-100≠140.

edit: BTW, this kind of mistake is incredibly common when doing this sort of problem. We ALL do it, all the time. You should just get used to checking for them. It's tedious, but the effort saved by avoiding confusion is worth it in the long run.

 

[PhysicsTest]

Yes, i got the equation as
$i_l = \frac 9 {100}(1- e^{-100t})$Amps
and
$v_l = 36e^{-100t}$
The time constant is 10ms.
At 25ms
$i_l=82.6 A$ $v_l = 2.95V$

But one question that i want to understand is the negative of the voltage source is not ground reference. As per the theory the ground is low resistance path so the current will try to flow to the ground or negative side of the battery?

 

[DaveE]

"Ground" can be thought of as an arbitrary choice for the name of a circuit node. The analysis of this schematic is the same regardless of where you define ground to be. When you were figuring out KVL around each of those loops, did ground matter for your resulting equations?

Ground may not always be "the low resistance path", it is usually just a name. Ground really can matter if you are dealing with circuits that interface with other circuits, then "ground" may be a shared node between circuits.

 

[phinds]

the current will try to flow to the ground or negative side of the battery?

In addition to @DaveE's explanation, keep in mind that current does not HAVE to flow to the negative side of a voltage source in such idealized problems. In real life you have issues in trying to swamp a battery or power supply with reverse current but in idealized beginner's problem, that is not the case. I'm NOT suggesting that it's happening in this case, I'm just pointing it out for completeness sake.