Capacitors - Alternative to practical

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#1
Could someone help me with 2fii) please?

http://www.edexcel.com/migrationdocu...e_20130115.pdf

In the MS it shows that the position of the resistor is simply swapped, but I don't understand how that would help to cut down the time delay, could someone explain it to me please?

Here's a link to the mark scheme: http://www.edexcel.com/migrationdocu...s_20130307.pdf

Thank you
0
6 years ago
#2
(Original post by jtbteddy)
Could someone help me with 2fii) please?

http://www.edexcel.com/migrationdocu...e_20130115.pdf

In the MS it shows that the position of the resistor is simply swapped, but I don't understand how that would help to cut down the time delay, could someone explain it to me please?

Here's a link to the mark scheme: http://www.edexcel.com/migrationdocu...s_20130307.pdf

Thank you
Ask yourself how long it's necessary to wait before switching from X to Y in the original circuit?

Spoiler:
Show
The capacitor must charge through the reisistor which is limiting the value of the current. The CR product (time constant) is circa 22 seconds, meaning the capacitor will take t = 5CR (>110 seconds) to get to within 99.5%of full charge.

With the modified circuit, there is no resistor (R <<1 ohms wire resistance) limiting the charging current and the capacitor can reach full charge virtually instantaneously after the switch moves from X to Y.
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#3
(Original post by uberteknik)
Ask yourself how long it's necessary to wait before switching from X to Y in the original circuit?

Spoiler:
Show
The capacitor must charge through the reisistor which is limiting the value of the current. The CR product (time constant) is circa 22 seconds, meaning the capacitor will take t = 5CR (>110 seconds) to get to within 99.5%of full charge.

With the modified circuit, there is no resistor (R <<1 ohms wire resistance) limiting the charging current and the capacitor can reach full charge virtually instantaneously after the switch moves from X to Y.
Okayy, i got it! Thank youu

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0
6 years ago
#4
Can we discuss unit 6b please. Where are the people.
0
6 years ago
#5
(Original post by Leil)
Can we discuss unit 6b please. Where are the people.
Do you have a specific question?

Thanks.
0
6 years ago
#6
(Original post by uberteknik)
Ask yourself how long it's necessary to wait before switching from X to Y in the original circuit?

Spoiler:
Show
The capacitor must charge through the reisistor which is limiting the value of the current. The CR product (time constant) is circa 22 seconds, meaning the capacitor will take t = 5CR (>110 seconds) to get to within 99.5%of full charge.

With the modified circuit, there is no resistor (R <<1 ohms wire resistance) limiting the charging current and the capacitor can reach full charge virtually instantaneously after the switch moves from X to Y.

why should I use the resistor to limit the value of current?
why can't I charge the capacitor without the resistor?
0
6 years ago
#7
(Original post by Lamalam)
why should I use the resistor to limit the value of current?
why can't I charge the capacitor without the resistor?
OK. This is where engineering and physics mesh together.

The raw theory and equations will not tell you why, simply that it's a possible and not prohibited by any fundamental physical laws.

The engineering of the components (capacitor construction, materials, cables, conections etc) have physical limitations and if stressed beyond those limits (like stretching a wire beyond it's elastic limit), permanent damage may and does occur.

For instance: if the current charging the capacitor is not limited, at switch on, the capacitor presents a short circuit to the supply. Theoretically, the 'inrush' current can briefly be infinite.

I = V/R and as R approches 0, I approaches infinity.

Power = V2/R. As R approaches zero, power tends to infinity.

That means all of the energy available in the power supply will try to be dumped onto the capacitor plates in zero time. The energy spike caused by that action may easily be enough to destroy the capacitor and anything else in the path of the current.

The bigger the capacitor and the lower the series resistance, then the greater the chance of that occuring. Indeed, it can lead to hazardous situations where electric shock, fire or even explosion is possible.

Large electrolytic and high capacitane tantalum devices are especally vulnerable.

This is why it is always advisable to check the safety implications of any experiment before you execute yourself and other with you!
0
6 years ago
#8
(Original post by uberteknik)
OK. This is where engineering and physics mesh together.

The raw theory and equations will not tell you why, simply that it's a possible and not prohibited by any fundamental physical laws.

The engineering of the components (capacitor construction, materials, cables, conections etc) have physical limitations and if stressed beyond those limits (like stretching a wire beyond it's elastic limit), permanent damage may and does occur.

For instance: if the current charging the capacitor is not limited, at switch on, the capacitor presents a short circuit to the supply. Theoretically, the 'inrush' current can briefly be infinite.

I = V/R and as R approches 0, I approaches infinity.

Power = V2/R. As R approaches zero, power tends to infinity.

That means all of the energy available in the power supply will try to be dumped onto the capacitor plates in zero time. The energy spike caused by that action may easily be enough to destroy the capacitor and anything else in the path of the current.

The bigger the capacitor and the lower the series resistance, then the greater the chance of that occuring. Indeed, it can lead to hazardous situations where electric shock, fire or even explosion is possible.

Large electrolytic and high capacitane tantalum devices are especally vulnerable.

This is why it is always advisable to check the safety implications of any experiment before you execute yourself and other with you!
"The bigger the capacitor ...." Big in the sense of large capacitance right?? and what is series resistance??

THANK YOUfor spending time to answer our questions in TSR
0
6 years ago
#9
(Original post by Lamalam)
"The bigger the capacitor ...." Big in the sense of large capacitance right??
Yes. The large surface area of the plates are able to hold a lot of electrons. i.e. energy.

(Original post by Lamalam)
and what is series resistance??
Any resistance in the path between the supply and the capacitor i.e. resistance in the path of the current charging the capacitor.
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