- #1
ZxcvbnM2000
- 64
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Homework Statement
The circuit shown in Figure 5 contains two emfs of opposite polarity E1 = 40 V,
and E2 = -20 V. The circuit also contains a switch, resistors R1 and R2, and a
capacitor C.The capacitor is initially discharged.
The switch is connected to position ‘a’ for a period of 8 seconds, after which it is
returned to the unconnected position ‘b‘ for a period of 12 seconds before being
connected to position ‘c’ for a further 8 seconds.
What are the time constants for each of the three time intervals?
Calculate the potential difference across the capacitor, VC, at the end of
each time interval.
[URL=http://imageshack.us/photo/my-images/651/dsdxsdxs.jpg/][PLAIN]http://img651.imageshack.us/img651/3112/dsdxsdxs.th.jpg[/URL][/PLAIN]
Homework Equations
The Attempt at a Solution
From 0-8 seconds the time constant is T=CR = 5.44 s
From 8-20 seconds T is 12 seconds
From 20- 28 seconds T is 5.44 s
Capacitor voltage at point A : Vc=40(1-e^(-8/5.44)) = 30.808 V ( Charging )
Voltage at B : ( Discharging) Vc' = 30.808e^(-12/12) = 11.33 V
Voltage at C : I don't know what to do . At first the capacitor has 11.33 V but because the polarity of this source is reversed then the capacitor will have to discharge ( go to 0 Volts ) . Is that right ? I am really confused here . Because these 8 seconds from 20-28 it will take some time to discharge from 11.33 to 0 and then recharge again with the other polarity .. WHAT DO I DO ?!
Thank you :)