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OhBoy
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Homework Statement
Having issues with the some signs when applying kirchhoffs law. When you take the derivative of charge with respect to time, is it always -i? I figured if it's charging, the change in charge with respect to time would be the same as just i, but if it's discharging then it would be -i? Am I correct here?
Also, when you're integrating current, when is it and isn't it -Q as opposed to just Q? For some reason, conceptually I can only understand summing up the current always being just positive Q... But I guess if it's discharging then the integral of current would be -Q and integrating charging current would be Q? Or, if discharging current, Q(t) = Q(0) - integral of I(t) dt? That would get me Q(0) - Q(t) = integral of I(t) dt, initial is larger than later in time when it loses charge but that would get me integrating current for discharging current is +Q not -Q.
Also, when does an inductor supply voltage? When it was 'charged' up by an emf source and that emf source is disconnected? If an inductor is hooked up to a resistor, does kirchhoffs law predict L di/dt = IR? Ldi/dt - IR = 0? I'm imagining an inductor and resistor in parallel, being charged up, and a switch opning disconnecting the battery from the inductor and resistor. The thing is, when I do that approach and do the math, I get I = I(0) * e^(tR/L), which predicts the current goes up, not down... What's going on here? I try to justify my initial set up because when switch is opened, current weakens, inductor as a response supplies an emf in direction of original current (meaning potential is gained?).
Homework Equations
V = IR + L di/dt + Q/C
L di/dt = IR ? L di/dt = -IR?
The Attempt at a Solution
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