Find current in circuit (DC RL circuit with switch)

[jaus tail]

Homework Statement

Initially switch is at 0.
At t = 0 seconds switch is put at 1.
After 1 time constant switch is put at 2.
Find equation for current for t > 2[/B]

Homework Equations

current through inductor tries to remain constant.
i(t) through inductor in series RL circuit is i(t) = [ i(initial) - i(final) ] e^(-R*t/L + i(final)

The Attempt at a Solution

At t < 0 I(t) is 0.
For t > 0 switch is in position 1.
i(t = 0⁺ ) = i(t = 0^- ) = 0.
for t much more than 0, the inductor will be short circuit and i will be 100 volts / 1 ohm = 100 A.

so equation for t > 0 = (initial current - final current) e^-R*t/L + final current
= (0 - 100)e^{(-1 * t / 1} + 100 = 100 (1 - e^(-t))

Now at t = 1 time constant switch is at position 2.
time constant = L/R = 1/1 = 1 second.
At t = 1 second, i(t) = 100 (1 - e^-1)
= 63A

Now at t = 1 second switch is at position 2.
So initial current is same as earlier which is 63A.
Final current is 50A.
So we get answer as [ initial - final ] e^(-t) + final
which is (63 - 50 ) e^-t + 50
= 50 + 13e^(-t)

But in answer they've given
50 + 13e^-(t-1)

Where did the -1 part come from?

 

[gneill]

Where did the -1 part come from?

Presumably they wish to keep the zero time as the instant the switch moved from position 0 to position 1. On that basis, what time does the switch move to position 2?

 

[jaus tail]

Switch moves to position 2 at 1 second.

 

[gneill]

Switch moves to position 2 at 1 second.

Correct. And when you use the general expressions to write the current for an LR circuit, what is the assumed starting time?

 

[jaus tail]

That's mostly written in question. Like we have to find current across L at t = 0⁺.
If question says. 'find current just after switch is closed' we write i(t⁺) = i(t^-)

 

[gneill]

That's mostly written in question. Like we have to find current across L at t = 0⁺.
If question says. 'find current just after switch is closed' we write i(t⁺) = i(t^-)

Forget the problem for a moment. What do the general equations assume for the time start?

 

[jaus tail]

Zero I guess... But why should answer have t - 1?
The whole system starts at same time of 0 second.

 

[gneill]

Zero I guess... But why should answer have t - 1?
The whole system starts at same time of 0 second.

The general equations assume t starts at zero for the curve they describe. When you use them to describe a curve at some other starting time, this does not change the t=0 assumption for the math. It's up to YOU to shift YOUR time base so that the math "thinks" they start at zero as they were designed to do.

 

[jaus tail]

Ok, so the switch at 2 is at t = 1 seconds, so that means that the 50 V circuit is activated at t = 1 second.
However my equations of
i (t) = ( initial - final ) e^-R*T/L start from T = 0 seconds.

But since t = 1(time when 50V is activated) , we have t - 1 = 0 and this = T(from equation above).
So we replace T by t - 1.
Like the origin is shifted ahead. Just like lag network.

 

[gneill]

Yes, that's right. You've probably dealt with similar situations when writing time shifted versions of the unit step or unit impulse functions.

 

[jaus tail]

yes I've solved questions with expressions u(t-1) or delta (t + 1) but didn't thought how it'd be in words. Thanks for the explanation.