# Capacitors circuit

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#1 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?

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5 years ago
#2
(Original post by jtbteddy) 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.
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#3
(Original post by Stonebridge)
The question says charges, not discharges.

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

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: 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?

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5 years ago
#4
(Original post by jtbteddy)

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: 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 ...
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#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
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.
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#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
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.
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#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
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#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?
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