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capacitor in a circuit, what exactly is happening? ? :(
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soo since the switch is joined to X, that means thr capacitor is charging, great. So the upper plate will be positively charged. What about the ammeter? will current be increasing in it or decreasing? and in what direction? (cause wouldn't the positive plate of the capacitor be a source of current to the ammeter, which will be opposite to the direction of the current from the cell) I don't think anyone will understand what I said haha, I don't think im very clear. But HELP WOULD BE GREATLY APPRECIATED!!![Name: 20171220_105526-compressed.jpg.jpeg
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#3
(Original post by sarah99630)
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when you start to consider capacitors (and inductors) you need to be very clear about when things happened and also what the previous condition was. These components store energy so they have a 'memory' of what's happened to them in the recent past.
Questions will typically give you a story explaining what has previously happened in a capacitor circuit and you need to follow and understand that to make sense of the question.
You'll get a transient condition the instant you close the switch and it'll settle down to a steady condition after the passage of time.
So the question 'what's happening in this circuit' doesn't have any meaning until you specify how long that switch has been in that position and what the charge on the capacitor was immediately before the switch was closed.
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(Original post by Joinedup)
With circuits made of just resistors and batteries it doesn't matter whether you close a switch a millisecond ago or an week ago... if you replace the capacitor in that circuit with a resistor it makes sense to ask 'what's happening in this circuit' without needing to know how long the switch has been closed or what the resistor was doing the instant before the switch closed.
when you start to consider capacitors (and inductors) you need to be very clear about when things happened and also what the previous condition was. These components store energy so they have a 'memory' of what's happened to them in the recent past.
Questions will typically give you a story explaining what has previously happened in a capacitor circuit and you need to follow and understand that to make sense of the question.
You'll get a transient condition the instant you close the switch and it'll settle down to a steady condition after the passage of time.
So the question 'what's happening in this circuit' doesn't have any meaning until you specify how long that switch has been in that position and what the charge on the capacitor was immediately before the switch was closed.
With circuits made of just resistors and batteries it doesn't matter whether you close a switch a millisecond ago or an week ago... if you replace the capacitor in that circuit with a resistor it makes sense to ask 'what's happening in this circuit' without needing to know how long the switch has been closed or what the resistor was doing the instant before the switch closed.
when you start to consider capacitors (and inductors) you need to be very clear about when things happened and also what the previous condition was. These components store energy so they have a 'memory' of what's happened to them in the recent past.
Questions will typically give you a story explaining what has previously happened in a capacitor circuit and you need to follow and understand that to make sense of the question.
You'll get a transient condition the instant you close the switch and it'll settle down to a steady condition after the passage of time.
So the question 'what's happening in this circuit' doesn't have any meaning until you specify how long that switch has been in that position and what the charge on the capacitor was immediately before the switch was closed.
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#5
(Original post by sarah99630)
soo since the switch is joined to X, that means thr capacitor is charging, great. So the upper plate will be positively charged. What about the ammeter? will current be increasing in it or decreasing? and in what direction? (cause wouldn't the positive plate of the capacitor be a source of current to the ammeter, which will be opposite to the direction of the current from the cell) I don't think anyone will understand what I said haha, I don't think im very clear. But HELP WOULD BE GREATLY APPRECIATED!!![Name: 20171220_105526-compressed.jpg.jpeg
Views: 45
Size: 32.3 KB]()
soo since the switch is joined to X, that means thr capacitor is charging, great. So the upper plate will be positively charged. What about the ammeter? will current be increasing in it or decreasing? and in what direction? (cause wouldn't the positive plate of the capacitor be a source of current to the ammeter, which will be opposite to the direction of the current from the cell) I don't think anyone will understand what I said haha, I don't think im very clear. But HELP WOULD BE GREATLY APPRECIATED!!
What rule/phenomenon can you use to work out the direction of the current?
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(Original post by CheeseIsVeg)
What equation relates current and capacitance?
What rule/phenomenon can you use to work out the direction of the current?
What equation relates current and capacitance?
What rule/phenomenon can you use to work out the direction of the current?
(Original post by CheeseIsVeg)
What equation relates current and capacitance?
What rule/phenomenon can you use to work out the direction of the current?
What equation relates current and capacitance?
What rule/phenomenon can you use to work out the direction of the current?
Thanks a lot for making me think with those questions
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#7
(Original post by sarah99630)
that question was later talking about the switch moving to Y then asked what's happening to the capacitor so the answer was : discharges exponentially. And then drawing the graph of current in the ammeter when that happens, which was easy, again. then ha!! the trouble started when they asked what would happen to the graph when the switched is RETURNED to X? and it was still a discharge graph but on the opposite side. So im doing some reading around that because clearly, looks like I have many misconceptions on it.
Thanks a lot for making me think with those questions
that question was later talking about the switch moving to Y then asked what's happening to the capacitor so the answer was : discharges exponentially. And then drawing the graph of current in the ammeter when that happens, which was easy, again. then ha!! the trouble started when they asked what would happen to the graph when the switched is RETURNED to X? and it was still a discharge graph but on the opposite side. So im doing some reading around that because clearly, looks like I have many misconceptions on it.
Thanks a lot for making me think with those questions
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