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I've got this problem:

enter image description here

So I'm thinking about the discharged section, t<0 s. Mainly my thoughts are how there could be a voltage across the capacitor when it is discharged. Basically I want to understand what happens when the capacitor gets discharged and charged. My own thoughts are that between -2 ms to 0 ms there is no current flowing because of the switch being open. Then there should be no charge "applied" on the capacitor. Therefore I'm thinking that the volatge over the capacitor should be 0 V between -2 ms and 0 ms.

Am I thinking right?

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If the capacitor is discharged, then there is no voltage across it, because that's what "discharged" means. So you are right; the voltage is zero until the switch is turned on.

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  • \$\begingroup\$ Correct me if I'm wrong, but between 0 ms and 2 ms, that will say t>0 when the switch opens at t = 0, I get the voltage to 2 V (linear) by using v(t) = I/C x t. Am I correct? \$\endgroup\$ – Vetenskap May 7 at 12:29
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    \$\begingroup\$ @Vetenskap; this is not a discussion site. You ask a question and we try to answer. You don't ask a question and then ask a different question as a comment to the answer. \$\endgroup\$ – Charles Cowie May 7 at 13:41
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Discharging a capacitor refers to removing a constant DC voltage that is removed from a capacitor by usually a low value resistor across it's two terminals. A capacitor only passes AC current and blocks DC current. A circuit may have both AC and DC current available. Turning off it's switch stops the AC component at the capacitor but not the DC component. That is why you need to discharge a capacitor that can hold a DC charge voltage if no discharge resistor is present.

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