[mediman]

An LR circuit contains a small resistance of r = 1 Ohm and inductance of 2mH in series with a battery of emf 2 volts. The switch is initially in a closed state. Then, at t = 0, the switch is opened so that the large circuit breaker resistance of Rcb= 2000 Ohms is now in series with the other circuit elements.
If the switch had been in the closed position for a long time before t = 0, could anyone tell me what was the steady state current flowing the circuit for t < 0?

Now, with the steady current flowing, the switch is opened at t = 0, so that the circuit resistance is now R = Rcb + r, I need to write a differential equation for the current as a function of time which describes the change in the current for t > 0. I then need to solve this equation (by integration) to determine the current as a function of time under the approximation that the battery emf = 0.
Finally, I have to find the potential drop across the circuit resistance of R = 2001 Ohms just as the switch is opened and calculate the elapsed time after the switch is opened at which the potential drop across the circuit resistance has reduced to 50 Volts. Thanks.

[harum]

If the switch had been in the closed position for a long time before t = 0, could anyone tell me what was the steady state current flowing the circuit for t < 0?

You have to be able to answer the first question: it only takes Ohm's Law for direct currents. Only after you have the answer for the steady state current for t < 0, you can move on, because the second question takes some knowledge about inductance.

What is your answer for the first question? Can you also provide the drawing of the circuit?

[mediman]

OK, I think that for the first question, the steady state current is just obtained by using Ohms Law, V = IR, so that if the battery emf is 2 volts and the resistance is 1 ohm, then, I(s) = V / R = 2 Amps.
I have studied a derivation for the equation to determine I(t) under the approximation that the battery emf = 0 and the equation says that:
I(t) = I(s) (1 - (e - (R/L))t) where e is an exponential.
This is where I get confused, because when the circuit is switched on, the resistance in the circuit goes from 1 ohm to 2001 ohms (in the circuit, there is a 2000 Ohm resistor across the switch (called the circuit breaker resistance) and it is in series with an ordinary resister of 1 Ohm and the inductor L is placed after the 1 Ohm resistor).
Would the voltage drop across the resistor just as the switch is turned on be = IR = (2 Amps)(2001 Ohms) = 4002 volts?? (this seems a lot).
and then, could you calculate the current when the voltage is 50 volts, using I = V / R = 50 / 4002 = 0.0125 Amps. Then, could you plug in these values into the above equation to work out t? As in, I(t) = 2 (1 - (e - (2001 / 2 x 10^(-3))t).
But then, I get a time of about 200,000 seconds!!

[harum]

Yes, this is the answer for the first question. The expression for I(t) that you have found describes the circuit for t < 0, i.e. when the current was growing from zero to I(steady), before the emf source was switched off. When you switch the emf source off, the cuicuit changes and you have to solve new (simpler) equation with new time coordinate, because you have just created a new circuit by using the switch and steady values of the first circuit become initial conditions for the second:

L(dI/dt) + RI = 0; here R = r + Rcb, and emf = 0.

Compare this to the circuit before switching the emf off:

L(dI/dt) + RI = E, here E is voltage of emf and R = r.

As for the initial drop, yes, it is this big. This is the purpose of such a load -- to turn all of the residual current into heat very quickly.

Solve the simpler DE and use it for the last question.

[mediman]

Yes, this is the answer for the first question. The expression for I(t) that you have found describes the circuit for t < 0, i.e. when the current was growing from zero to I(steady), before the emf source was switched off. When you switch the emf source off, the cuicuit changes and you have to solve new (simpler) equation with new time coordinate, because you have just created a new circuit by using the switch and steady values of the first circuit become initial conditions for the second:

L(dI/dt) + RI = 0; here R = r + Rcb, and emf = 0.

Compare this to the circuit before switching the emf off:

L(dI/dt) + RI = E, here E is voltage of emf and R = r.

As for the initial drop, yes, it is this big. This is the purpose of such a load -- to turn all of the residual current into heat very quickly.

Solve the simpler DE and use it for the last question.

OK, thanks, makes a lot more sense now.
So, knowing that the Steady State current is 2 Amps and the potential drop across the circuit breaker resistance is 4002 Volts, I tried to solve the simple Differential Equation and got the following:
I(t) = (emf / R) (e - (R/L) t ), where e - (minus) is the negative exponential.
I presume then that emf / R is just the steady state value of the current (2 amps?? )
However, I am not sure if I have done the right calculation to find the time taken for the voltage drop across the circuit breaker resistance to fall to 50 volts.
If I use R = r + Rcb, I get 2001 Ohms and if I divide this by L above (which has a value of 2 x 10^(-3) H, I get a negative exponential of e - 1000,500t. This still gives me a very large time for the voltage drop to go to 50 volts.
Also, I am not sure if I am using the right values for I(t) in the equation or how do I take the 50 volt value into account in the above equation.

[harum]

I presume then that emf / R is just the steady state value of the current (2 amps?? )

Not sure how you can ask this question if you are saying you have solved the simpler equation...

Can you show how did you get from here: L(dI/dt) + RI = 0
to here ?:

I(t) = (emf / R) (e - (R/L) t )

Where does emf in your answer come from?

It is always a good thing to check if answers in physics problems are realistic, just like you do. If they are not, then either equations or calculations are wrong. I suspect your calculations are off. Can you show the formula for calculating time and then the same formula with the numbers plugged?

Do you understand that your current will be dropping exponentially very quickly starting at time t = 0 from its steady state value towards zero? Initial drop is (Rcb+r)*I(s), the drop at all other times is (Rcb+r)*I(t). You have found I(t), meaning that you can calculate I at any time as well as you can calculate how long it takes to go from I(s) to any lower value.

[mediman]

Not sure how you can ask this question if you are saying you have solved the simpler equation...

Can you show how did you get from here: L(dI/dt) + RI = 0
to here ?:

Where does emf in your answer come from?

It is always a good thing to check if answers in physics problems are realistic, just like you do. If they are not, then either equations or calculations are wrong. I suspect your calculations are off. Can you show the formula for calculating time and then the same formula with the numbers plugged?

Do you understand that your current will be dropping exponentially very quickly starting at time t = 0 from its steady state value towards zero? Initial drop is (Rcb+r)*I(s), the drop at all other times is (Rcb+r)*I(t). You have found I(t), meaning that you can calculate I at any time as well as you can calculate how long it takes to go from I(s) to any lower value.

OK, sorry, I'm not completely getting this. And, I think you are correct about the emf - it shouldn't have been in my answer. So, to try and make it brief, I solved the differential equation the following way:
from L(dI/dt) + RI = 0, I got di / -i = R / L dt
Then I integrated this and got ln (-i) = (R / L) t + C
Then, I set the initial conditions to find C using: at t = 0, I = I(s), which is the steady state current, so that C = ln (-I (s))
This eventually left me with ln i = ln I(s) - (R / L) t
and multiplying across by the exponential to cancel the ln terms, I got:
i(t) = I(s) e - (R / L) t
So then, I need to work out the time (t) when the voltage drop across the circuit breaker resistance drops to 50 Volts.
If I = V / R, then when V drop = 4002 volts, I = 4002 / 2001 = 2 Amps (the steady state value).
Then when V drop = 50 volts, I = 50 / 2001 = 0.025 Amps (approx.)
I then used this value as I(t) in the final equation above, so that
0.025 = 2 (e - (2001 / 0.002) t
where R = Rcb + r = 2001 Ohms and the inductance, L = 0.002 H.
then, taking 2 over to the left hand side, I get:
0.0125 = e - (2001 / 0.002) t,
Then, taking the ln on both sides to get rid of the exponential,
I get : - 4.38 = - 1000,500 t,
which gives me a time of over 200,000 seconds, which is way too high, considering that the current drops very quickly, as you pointed out.

[harum]

I get : - 4.38 = - 1000,500 t,
which gives me a time of over 200,000 seconds, which is way too high, considering that the current drops very quickly, as you pointed out.

emf should be in I(t) because the steady state current I(s) is emf/r. My question was, how did you put the value for emf into I(t) without knowing that it was done only to explicitly express I(s) in known terms.

You are one correct calculation away from the correct answer.
Look at it again:

- 4.38 = - 1000,500 t

I would suggest to only plug numbers in the final expression for the answer, i.e. for this problem:

t = ln ( (V(50)*r)/(emf*(Rcb+r)) ) * (L/(Rcb+r));

then you check your answer for extreme values and dimensions and plug numbers, if satisfied.

[mediman]

emf should be in I(t) because the steady state current I(s) is emf/r. My question was, how did you put the value for emf into I(t) without knowing that it was done only to explicitly express I(s) in known terms.

You are one correct calculation away from the correct answer.
Look at it again:

- 4.38 = - 1000,500 t

I would suggest to only plug numbers in the final expression for the answer, i.e., for this problem:

t = ln ( (V(50)*r)/(emf*(Rcb+r)) ) * (L/(Rcb+r));

then you check your answer for extreme values and dimensions and plug numbers, if satisfied.

That was a silly maths mistake I made. Thank you for pointing it out.
Dividing 1000,500 into 4.38, I get 4.38 microseconds.
And yes, the final expression you provided for t is helpful for understanding it.
Thanks again.

[harum]

Except that I also made a mistake, lost the minus sign, and the answer should be:

t = - ln ( (V(50)*r)/(emf*(Rcb+r)) ) * (L/(Rcb+r)), or

t = ln ( (emf*(Rcb+r)/(V(50)*r)) ) * (L/(Rcb+r)).

Again, this mistake was obvious because the higher the emf is, the longer time t is; and the higher the desired voltage drop, the shorter the t.

So, make sure your numeric answer is accurate.