- #1
It means that all circuit will have only resistors?Ok.MisterX said:I assume you should assume that the circuit is in steady state before the switch is closed.
If that is the case, no current will flow through the capacitor, and the inductor will be like a short circuit.
MisterX said:You could use the Laplace transform of the circuit after the switch is closed to get the transients.
Is that 8 a typo, or is it really 8*K2 ? These problems with the "K" constants seem a bit odd. Why not just specify the component values and be done with it? Also, are we to take the resistance values to have implied Ohms units? (the capacitors and inductors have specified millihenries and microfarads). The current, too, lacks explicit specification for units. Do we assume Amps and not milliamps?builder_user said:The Attempt at a Solution
J=4*K1
L1=1mH*K2
C1=10mkF8*K2
:
:
MisterX said:Here's how:
Replace the circuit elements with their http://en.wikipedia.org/wiki/Laplace_transform#s-Domain_equivalent_circuits_and_impedances". Solve for the Laplace transform of the variables of interest. Then find the inverse Laplace transform of those variables. Alternatively, you could write out the differential equations.
gneill said:Is that 8 a typo, or is it really 8*K2 ? These problems with the "K" constants seem a bit odd. Why not just specify the component values and be done with it? Also, are we to take the resistance values to have implied Ohms units? (the capacitors and inductors have specified millihenries and microfarads). The current, too, lacks explicit specification for units. Do we assume Amps and not milliamps?
builder_user said:It's the last task.
What's happen with capacitor?
I need to find Thevenin equivavalent before and after commutation right?
gneill said:The capacitor will have some initial voltage on it when the switch is closed. The inductor will be carrying some initial current.
Are you required to find U(t) and i(t) for the inductor in the problem, or just the initial values after the switch closes? Otherwise, since there's both L and C in the circuit there will probably be some oscillation after the switch closes; you might have to deal with damped oscillation (second order differential equation rather than first order).
You can probably skip finding the Thevenin equivalent for the 'before' circuit if you can determine the voltage on the capacitor and current through the inductor before the switch is closed.
gneill said:The "after" circuit looks fine. Can you tell the capacitor voltage and inductor current from the first circuit?
builder_user said:I...know it?Or I need to find it?
gneill said:You should be able to calculate the steady state value of the voltage on the capacitor, and the steady state current flowing through the inductor before the switch is closed.
gneill said:Yes...
builder_user said:Before..?
gneill said:i2 = J*R3/(R1 + R2 + R3)
gneill said:That's right. It's a current divider.
Draw a circuit with a current supply I driving two parallel resistors, Ra and Rb. How does the current divide between the Ra path and the Rb path?
The parallel resistance is Ra*Rb/(Ra+Rb). So the voltage across the pair is V = I*Ra*Rb/(Ra+Rb). Thus, the current through, say, Ra, is V/Ra, which is I*Rb/(Ra+Rb).
Now go back to your circuit. One path has resistance R1+R2, while the other has resistance R3.
gneill said:Before proceeding, let's make an educated guess at what the voltage across the inductor is going to look like when the switch closes. That way we'll know where we're headed.
When the switch closes you'll have the current supply J = 12A in parallel with R2 and C1 and the branch R3&&L1 (using your && notation). The capacitor has an initial voltage of UC = UR2 = 9.6V (correct?). The inductor is carrying initial current iL = 9.6A (correct?).
The instant the switch closes, the capacitor will want to maintain its voltage and the inductor will want to maintain its current. So the top of R3 will be at potential UC = 9.6V. The current through R3, however, will still be iL = 9.6A, so the voltage drop on R3 is going to be 19.2V. That means the voltage across L1 must shoot up to 9.6V (- on top, + on bottom of L1) in order to satisfy Kircchoff, because UR3 + UL1 = UC at that instant.
Eventually the voltage across L1 must return to zero in a new steady state. So there will be an exponential decay (with time constant to be determined) from 9.6V down to zero. Superimposed on that will be whatever oscillations, if any, that result from the interaction of L and C. But the "big picture" is that UL1 will shoot up to 9.6V and decay.
Does all that make sense?
gneill said:Isn't J = 4*K1 = 4*3 = 12? Or are the units something other than amps?