Thanks for showing your work and making an honest attempt.
Your solution for t=0+ is correct.
A voltage rate-of-change of 1 mV/s is NOT a voltage of 1 mV. Track your units and you would have seen that they don't work out.
You are actually almost tracking your units, but you aren't paying attention to them.
You have V = L (di/dt) yet the values you throw in this equation have units of L (mH) and current (A), not L and di/dt (A/s). Had you tracked the units, you would have seen that they work out to volt-seconds. They should work out to volts, but you tacked on the units of volts/second for some reason.
Track your units! Most mistakes you make will mess up the units allowing you to detect and track down the error very quickly.
Also, always ask if the answer makes sense? Given your answer, how much power is being dissipated in the resistor at t=∞? Where is that energy coming from?
and i am sure you also know for the capacitor that:
i=C*dv/dt
and solving for dv we get:
dv=i*dt/C
Now looking at the time frame for the switch opening, how much time is there from:
t1=0+
to an infinitesimally small time (deltaT) later:
t2=t1+deltaT
?
You might think of deltaT as being equal to 1/a where 'a' approaches infinity.
That is the key to solving this. Figure out how much that difference is, then apply it to the two equations for L and C (you almost got the answer already).
I want to find out if I say. The voltage across the Inductor will be V= L.(di/dt) which will be 10mh x(0.1A)/dt since I calculated the current to be 0.1A across the Inductor. this will make the Voltage to be 1mV across the Inductor for a certain amount of time. hence the Inductor is in parallel to the capacitor. The Voltage of the Capacitor will be also 1mV and because the 50ohm resistor is in parallel to the capacitor the voltage will be the same so the current across the 50ohm resistor will be 1mV / 50ohm which will be 0.00002A (across the resistor for a certain amount of time)
Is my deduction correct?
Or will I say that the current across the capacitor is I= C.(dv/dt) which will be 1mF x(1mV/dt) making the answer be 1x10^-6A through the capacitor hence to find the current across the 50ohm resistor will simply take 0.1A - 1x10^-6
which will be 0.099999 Amps through the resistor?
Well it takes time for the voltage across the capacitor to build up to even a small voltage like 1uv given even a large current like 10 amps. It takes more time than 1/a (as a approaches infinity as before).
The time between t=0 and t=0+ is 1/a, which is an unmeasureably small increment of time, so how much could the voltage across the capacitor change given no time at all and your calculated current at t=0 of 0.1 amps?
But you already solved that apparently, and now you need to look for the solution at t approaching infinity.
So when the switch is first opened, the capacitor voltage is zero (as you know), then starts to build up over time because the inductor is pumping current into the capacitor now and the resistor is 50 ohms which only conducts some of the current, but not all.
Some time later, the voltage will reach some peak value because as the inductor looses energy into the resistor there is less and less available to charge the cap with. If the circuit oscillates, the capacitor voltage will become lower, then higher, then lower again, etc., until all the energy is taken up by the resistor. If the circuit does not oscillate, then the energy is taken by the resistor but the cap simply reaches a peak value and then starts to decrease again.
What do you think happens to the voltage across the cap and current though the inductor when either of these conditions comes into play after some very long time interval (we call t=infinity) ? If there is current in the inductor (even 1ua) or voltage across the cap (even 1uv) where could that energy come from, because if there is any voltage across the cap the 50 ohm resistor tries to discharge the cap, right?
In the more general case you might have to figure out the damping of the circuit so that you can determine if it oscillates or just damps out.
See if you can now figure out what happens at t=infinity.
Try to keep in mind these two solutions again we found earlier:
di=v*dt/L
dv=i*dt/C
and these basically tell us that:
1. the current through an inductor can not change in zero time, and
2. the voltage across the capacitor can not change in zero time.
So the changes can only occur slowly (relatively speaking) and can never occur instantaneously.
Thank you Sir,
From my deduction and from what you have discussed. as the circuit is open after t=0+ time and time approaches infinity the Current across the resistor will decrease based on the fact as you said, the capacitor will discharge and from my assumption the current should also decrease and become zero over the 50ohm resistor.
I want to find out if I say. The voltage across the Inductor will be V= L.(di/dt) which will be 10mh x(0.1A)/dt since I calculated the current to be 0.1A across the Inductor.
(di/dt) is the instantaneous rate at which the current is changing. In general, this quantity is different at every moment in time. Your 0.1 A is the total current in the inductor at a particular moment in time. The 'i' in that di/dt is i(t), namely a function of time. di/dt is merely the derivative of i(t) and, hence, is also a function of time.
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