jtbteddy
Badges: 5
Rep:
?
#1
Report Thread starter 5 years ago
#1
Name:  ImageUploadedByStudent Room1401048788.562612.jpg
Views: 549
Size:  54.8 KB

This is a screenshot from the examiners report.

The question was:

"A student is investigating how the potential difference across a capacitor varies with time as the capacitor is charging. He uses a 100microF capacitor, a 5V d.c. Supply, a resistor, a voltmeter and a switch.

Draw a diagram of the circuit he should use"

Why is the circuit in the picture correct? Shouldnt the resistor be connected down, so that when the switch is opened, the capacitor discharges through the resistor?


Posted from TSR Mobile
0
reply
Stonebridge
Badges: 12
Rep:
?
#2
Report 5 years ago
#2
(Original post by jtbteddy)
Name:  ImageUploadedByStudent Room1401048788.562612.jpg
Views: 549
Size:  54.8 KB

This is a screenshot from the examiners report.

The question was:

"A student is investigating how the potential difference across a capacitor varies with time as the capacitor is charging. He uses a 100microF capacitor, a 5V d.c. Supply, a resistor, a voltmeter and a switch.

Draw a diagram of the circuit he should use"

Why is the circuit in the picture correct? Shouldnt the resistor be connected down, so that when the switch is opened, the capacitor discharges through the resistor?


Posted from TSR Mobile

The question says charges, not discharges.

When the switch is closed the cell charges the capacitor through the resistor.
0
reply
jtbteddy
Badges: 5
Rep:
?
#3
Report Thread starter 5 years ago
#3
(Original post by Stonebridge)
The question says charges, not discharges.

When the switch is closed the cell charges the capacitor through the resistor.
Got it, i misread it.

Could you also help me with this one, please?

http://pastpapers.edexcel.com/conten...e_20130613.pdf

Question 14cii)

Here's the mark scheme for it: Name:  ImageUploadedByStudent Room1401049698.932213.jpg
Views: 350
Size:  85.9 KB

I understand why a tangent has to be drawn, i got 2400Vs-1, but i dont understand how to get from there to find the current. Can you help please?


Posted from TSR Mobile
0
reply
Stonebridge
Badges: 12
Rep:
?
#4
Report 5 years ago
#4
(Original post by jtbteddy)
Got it, i misread it.

Could you also help me with this one, please?

http://pastpapers.edexcel.com/conten...e_20130613.pdf

Question 14cii)

Here's the mark scheme for it: Name:  ImageUploadedByStudent Room1401049698.932213.jpg
Views: 350
Size:  85.9 KB

I understand why a tangent has to be drawn, i got 2400Vs-1, but i dont understand how to get from there to find the current. Can you help please?


Posted from TSR Mobile

The charge on the capacitor at any time is

Q=CV

So if you have the rate of change of V from the tangent (this is dV/dt)

You have the rate of change of charge from dQ/dt = C . (dV/dt)

In other words, the rate of change of pd times the capacitance gives the rate of change of charge on the capacitor.
Rate of change of charge is the rate at which the charge is flowing off the capacitor.
And rate of flow of charge is ...
0
reply
jtbteddy
Badges: 5
Rep:
?
#5
Report Thread starter 5 years ago
#5
(Original post by Stonebridge)
The charge on the capacitor at any time is

Q=CV

So if you have the rate of change of V from the tangent (this is dV/dt)

You have the rate of change of charge from dQ/dt = C . (dV/dt)

In other words, the rate of change of pd times the capacitance gives the rate of change of charge on the capacitor.
Rate of change of charge is the rate at which the charge is flowing off the capacitor.
And rate of flow of charge is ...
I don't understand the bold bit .. Q = It, so dQ/dt should give me current?
0
reply
Stonebridge
Badges: 12
Rep:
?
#6
Report 5 years ago
#6
(Original post by jtbteddy)
I don't understand the bold bit .. Q = It, so dQ/dt should give me current?
Yes.
And you get dQ/dt by multiplying dV/dt by the capacitance C because Q=CV

You got dV/dt from the graph.
0
reply
jtbteddy
Badges: 5
Rep:
?
#7
Report Thread starter 5 years ago
#7
(Original post by Stonebridge)
Yes.
And you get dQ/dt by multiplying dV/dt by the capacitance C because Q=CV

You got dV/dt from the graph.
Looking at that now makes sense ...
The graph gives me dV/dt, and since Q = CV, C = dQ/dV

Multiplying dV/dt by dQ/dV cancels outs the dVs, giving me dQ/dt

Makes sense looking at it like that .. But how would I have thought of that in the exam? Can you help please? It wouldn't have occured to me
0
reply
Stonebridge
Badges: 12
Rep:
?
#8
Report 5 years ago
#8
(Original post by jtbteddy)
Looking at that now makes sense ...
The graph gives me dV/dt, and since Q = CV, C = dQ/dV

Multiplying dV/dt by dQ/dV cancels outs the dVs, giving me dQ/dt

Makes sense looking at it like that .. But how would I have thought of that in the exam? Can you help please? It wouldn't have occured to me
It comes with practice and experience.
You've seen it now, so next time it comes up (or something similar) you'll remember this example and hopefully make the connection.
0
reply
jtbteddy
Badges: 5
Rep:
?
#9
Report Thread starter 5 years ago
#9
(Original post by Stonebridge)
It comes with practice and experience.
You've seen it now, so next time it comes up (or something similar) you'll remember this example and hopefully make the connection.

Thank you
0
reply
jtbteddy
Badges: 5
Rep:
?
#10
Report Thread starter 5 years ago
#10
(Original post by Stonebridge)
The question says charges, not discharges.

When the switch is closed the cell charges the capacitor through the resistor.

Stonebridge, can you have a look at the graph again? If it's connected in parallel across the cell and also the capacitor, which PD is it actually measuring?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Regarding Ofqual's most recent update, do you think you will be given a fair grade this summer?

Yes (126)
29.44%
No (302)
70.56%

Watched Threads

View All