# Find current in circuit (DC RL circuit with switch)

1. Feb 2, 2017

### jaus tail

1. The problem statement, all variables and given/known data
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

2. Relevant 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)

3. 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)

50 + 13e-(t-1)

Where did the -1 part come from?

2. Feb 2, 2017

### Staff: Mentor

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?

3. Feb 2, 2017

### jaus tail

Switch moves to position 2 at 1 second.

4. Feb 2, 2017

### Staff: Mentor

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

5. Feb 2, 2017

### 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-)

6. Feb 2, 2017

### Staff: Mentor

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

7. Feb 2, 2017

### jaus tail

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

8. Feb 2, 2017

### Staff: Mentor

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.

9. Feb 2, 2017

### 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.

10. Feb 2, 2017

### Staff: Mentor

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

11. Feb 2, 2017

### 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.