- #1
marstery
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The problem is in bold and my logic/answers/questions are below. Please check my logic and any help with my uncertainties would be...well, helpful. thx
The switch has been in position 1 for a long time. Determine the voltage drop Vc(0) and V0(0-).
Since it has been a long time the capacitor acts as an open circuit so the current through the capacitor is 0ma and the current around the outer window is 12v/4k=3ma so V0(0-) is 3v. Current Ic is zero but was flowing to the right (contrary to the arrow drawn). Vc(0) is 12v since it has been a long time and the capacitor has charged fully.
The switch is thrown to position two at t=0. What is V0(0+)?
At t=0 the capacitor acts as a short and discharges in the direction that Ic is drawn. I3 is now 4ma. Since the cap is a short does all of the current flow into the cap so that V0(0+) is 0v for that moment?
Find the time constant and V0(t>5tc).
t=RC=3k*0.1 [tex]\mu[\tex]f=300[tex]\mu[\tex] s and after 5tc the capacitor is once again fully charged so all of the current is going the R1 and R2 so that V0(t>5tc) is 3v again.<br /> <br /> <b>write an expression for V0(t>=0)</b><br /> <br /> it will look like a logarithmic curve starting at 0 and leveling at (3 or -3v??) after 5tc, but i don't really know the eq for it.[/tex][/tex]
The switch has been in position 1 for a long time. Determine the voltage drop Vc(0) and V0(0-).
Since it has been a long time the capacitor acts as an open circuit so the current through the capacitor is 0ma and the current around the outer window is 12v/4k=3ma so V0(0-) is 3v. Current Ic is zero but was flowing to the right (contrary to the arrow drawn). Vc(0) is 12v since it has been a long time and the capacitor has charged fully.
The switch is thrown to position two at t=0. What is V0(0+)?
At t=0 the capacitor acts as a short and discharges in the direction that Ic is drawn. I3 is now 4ma. Since the cap is a short does all of the current flow into the cap so that V0(0+) is 0v for that moment?
Find the time constant and V0(t>5tc).
t=RC=3k*0.1 [tex]\mu[\tex]f=300[tex]\mu[\tex] s and after 5tc the capacitor is once again fully charged so all of the current is going the R1 and R2 so that V0(t>5tc) is 3v again.<br /> <br /> <b>write an expression for V0(t>=0)</b><br /> <br /> it will look like a logarithmic curve starting at 0 and leveling at (3 or -3v??) after 5tc, but i don't really know the eq for it.[/tex][/tex]
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