Oxford PAT 2012, Question 18

12 thoughts on “Oxford PAT 2012, Question 18”

1. ian says:

Hi, I am sorry but I think I havent been thought about the formula RC thing how come I= e^(t/RC)?

2. Jemisha says:

Hi,

Could you explain how you drew the second graph? I thought current was not a vector, so direction shouldn’t matter. Either way, when charging of the capacitor occurs, the current through the resistor eventually decreases (exponential decay) and then when discharging of the capacitor occurs through the resistor, the current eventually exponentially decays too? I am learning a lot of the PAT syllabus by myself and when I came across capacitor curves in my textbook and a number of websites, both the current-time graphs for charge and discharge have looked the same. Would you not gain any marks for drawing the same curves? And what is the time interval after no significant change is observed? I differentiated the function, and to work out the time I got -[ln(I) -ln(I initial)] RC = T. I can substitute values for RC and I initial in but I still have two unknowns. Could you explain how you got your value?

Thanks!

1. Jemisha says:

Also I forgot to add, I can understand why for charging curves the pd and charge across the the capacitor are initially zero explaining why the curve tends towards y = pd initial or charge initial, and why it is the reverse for discharge curves. However, for the current shouldn’t both graphs be the same?

Thanks! (apologies for the length of my question!)

1. You’re right that current is not a vector quantity, but it does still have a sign.

For example, think about a circuit with no capacitor, but just a resistor, R, with a battery across it that creates a voltage V across the resistor. We know the current I = V/R. Now swap the terminals on the battery so that the voltage is reversed: what was previously V is now -V. So the current is now -V/R, i.e. it is reversed too.

In the capacitor example, the current is originally flowing in the direction needed to charge up the capacitor, and subsequently flowing in the direction needed to discharge the capacitor. These directions are opposite, so the value of the second current is the negation of the first.

1. Jemisha says:

Okay, thanks, that makes sense. Is it correct to assume that conventional current is not observed in capacitor circuits? So the current does flow from the negative terminal to the positive. That’s the only way I think the graphs would make sense for current.

1. I’m not sure what “conventional current is not observed” means.

Let’s consider the details of what is happening here. Current is just the rate of flow of charge. ‘Conventional current’ is the rate of flow of positive charge, which is simply the reverse of the rate of flow of electrons (because electrons have negative charge), but that doesn’t really matter.

When a capacitor charges up it accumulates a positive charge on one plate and a negative charge on the other, and the more charge it accumulates, the voltage across ii increases (given by the equation Q = VC). What is actually happening as the capacitor is charging up is that electrons are flowing towards the negatively-charged plate and away from the positively charged plate, so there is a conventional current in the opposite direction, towards the positively charged plate.

When the capacitor discharges, the electrons are flowing away from the negatively-charged plate and towards the positively charged plate, so there is a conventional current in the opposite direction, away from the positively charged plate.

3. rickardo says:

I think there is a mistake in the first graph? Isnt the first graph supposed to be capictor discharge? When it is connected to d the capictor gets charged not discharged am i correct? So first one should be a charge graph and second discharge?

1. I don’t think there is a mistake in the first graph. Initially the switch is connected to e, and this charges the capacitor, and then after some time the switch is connected to d, and this discharges the capacitor. So the first graph is a charge graph and the second is a discharge graph.

So, why does the charge graph look like it does? When the capacitor is charging, it is in series with the resistor and the battery; the voltage across it is initially zero, so all the voltage is across the resistor, so the current is at its maximum. As the capacitor charges the voltage across it increases, so the voltage across the resistor decreases, so the current decreases …

So why does the discharge graph look like it does? When the capacitor is discharging, it is just in series with a resistor, and the battery has been switched out of the circuit. At the start of the discharge, the voltage across the capacitor is 12V, and it decays exponentially, and therefore so does the current. The current is flowing in the opposite direction from the charging current.

4. rickardo says:

How do you estimate the time in the “e” equation without a calculator /:

5. jon says:

where does the value of 30e-3 seconds come from? (no significant change) thanks!

1. Good question. A ‘significant change’ obviously a subjective thing. In this case, we know that dI/dt is 3(e^(-t/k))/4, where k is 8e-3, so I thought we would use a value for t of something over 3k, which is about 5 half lives.

It occurs to me that I haven’t strictly followed the instructions at the end of the question where it says to ‘indicate on your sketch the time interval T’. But I think it should be clear that the value of k is the same in each case.