**broken link removed** Originally Posted by
MrAl **broken link removed**
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We know the cap voltage is 10 volts and that cant change in zero time, so that makes v=-10 volts (it's negative because of the orientation in the circuit).
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Huh? Look at V(c) green line in the plot. I(R1) = I(L1) = blue line in the plot.
Hello Mike,
You'll note that you took that line out of context just a little, as others have noted when we solve for the current in the inductor we have to stay consistent with the current direction that was originally going through the inductor at some time before t=0. That would have made the current through the inductor oriented clockwise in that schematic (as usual we use conventional current flow, positive to negative). The lower case 'v' is the voltage with its polarity the way it would appear just after the switch is closed. This is overdamped, but it would work the very same way in the underdamped case if we made the cap value equal to say 0.005 F instead of 0.5 F.
If we didnt stay consistent in this way then the circuit would not work properly if we opened and closed the switch repeatedly as in PWM.
Using the definition for the inductor:
v=L*di/dt
that makes v=-10 volts.
If you still dont believe this, try modulating the switch using PWM at say 50 percent duty cycle and see what result you get.
Alternately we can assume a different current direction, but if we do chose to do that then we have to stay consistent with that direction too. That would make the capacitor voltage be negative at the start, and it would in fact stay negative, but for the purposes of evaluating the current v becomes the opposite polarity.
Maybe that is what caused the confusion. It's not that Vc changes itself, it's just that the polarity of v has to change for the purpose of evaluating the current through the inductor after the switch has just been closed (in the equation v=L*di/dt). Vc stays positive using conventional current flow.
Its good to discuss this because polarity in circuits like these is very important.
At the very least we should stay consistent with the direction of the current. If we start out clockwise, then we should finish clockwise or else note that we switched direction. Note that initially with the 20v source the current is oriented clockwise until it falls to zero, but then even a tiny leakage current in the cap 1e-32 amps would keep it clockwise. Thus it calculates out clockwise and then changes polarity, but you could say that it kept its polarity but changed direction if you really wanted to, as a positive current in the positive direction is the same as a negative current in the negative direction, and a positive current in the negative direction is the same as a negative current in the positive direction.
Interesting, if we look at the circuit in a circuit analyzer and allow time for the 20v source to pump up the cap, the current is positive in the resistor (and inductor) until the cap charges up, then when the switch is closed the current will dip negative. Try it.