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dddd12349999
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What is the equation switched capacitors
- Thread starter dddd12349999
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AI Thread Summary
The discussion centers on the equation relating charging current i(vdd) to a series of switched capacitors (c1 to c5) with sequential switch closures and a delay time (td). A critical issue is raised regarding the potential for infinite current when the second capacitor is switched, although this is noted to occur only momentarily. When the switch for C2 closes, the voltages across C1 and C2 equalize almost instantaneously, allowing both capacitors to continue charging. The conversation highlights the need for careful consideration of circuit behavior during the switching process. Understanding these dynamics is essential for effective circuit design involving switched capacitors.
- #2
Carl Pugh
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There is a problem with your circuit.
When the second capacitor is switched, there is an infinite current through the switch.
When the second capacitor is switched, there is an infinite current through the switch.
- #3
dddd12349999
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can we think about it as a variable capacitor?
- #4
SammyS
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dddd12349999 said:
That's only for an instant.Carl Pugh said:
When the switch for C2 is closed, the voltages across C1 & C2 become equal to each other virtually instantaneously. Then the pair continue to be charged for that point.
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
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In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question.
Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
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Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
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