Initially, the capacitor is charged with the charge of, lets say, 10 Coulombs, and the gate is open.
Let's make the resistance 100 Ohms and the capacitance 0.1 Coulombs/volt*. (aside: how are these values relevant to making part the less difficult than it otherwise might be?)
a) In this configuration, with the capacitor charged and the gate open, what is the voltage difference Va - Vb and why? (The “why” is just as important as the answer regarding Va - Vb.)
b) Now suppose you close the gate at t = 0. Think about what happens and answer the following question: roughly (within about 10% accuracy) what is the amount of heat energy generated in the resistor in 10 ms immediately after the gate is closed? ( Let's assume the resistor is a pure heater and that all the energy dissipated or converted ( or whatever term you want to use) in the resistor goes to heat.
c) Explain your thought process and how you obtained your result for part b). Explain why you think it's accuracy is good enough. Did you have to do any integration? Why or why not? (you can ask for a hint here. (via a comment))
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New problem about the same circuit: With reference to the same circuit, in which the capacitor is initially charged and the gate is closed at t=0.
a) graph Va-Vb as a function of time.
b) Graph Vb -Vc as a function of time.
c) What is the relationship between them?
* 0.1 Coulombs/Volt is an unrealistically large value for an ordinary capacitor. We use it here because otherwise the timescales would get very short, and having large negative exponents associated with the time can be distracting when one is first learning concepts.
PPS. I will avoid using farads for capacitance. One farad is the same thing as one coulomb per volt, so I don't think we need to introduce one more new unit.
is there anyone who can explain part b of the first problem? I used the equation U=(Q^2)/2C, but I am not sure if this is right. Does anyone understand this problem?
ReplyDeleteWhat about using R I ?
ReplyDeleteIgnore that. I don't know what i was thinking.
ReplyDeleteAnyway, part b asks about the resistor, yes?
"what is the amount of heat energy generated in the resistor in 10 ms"
So why not think about the resistor. Is R I^2 the energy generated* in the resistor per second? I think R I^2 is in Joules/second. SO in 10 ms, how many joules would be generated?
* One could say "generated" or "dissipated". They refer to the same thing from a different perspective. Basically energy being converted to heat...
So questions on the midterm are going to be this difficult? I really don't understand how to graph Va-Vb or Vb-Vc as a function of time at all. I appreciate any help that anyone can offer on this!
ReplyDeletewe did that on Friday. Would someone explain? Thanks.
ReplyDeleteeinino: can you graph e{-x}
ReplyDeleteAhh, unfortunately I have been very sick for the past week and I was unable to attend lecture on Friday. Is there a place in the book where I can learn how to do this on my own? I was looking at page 428 and that looked like it might be similar to what you're talking about, if I understand that section will I be okay with the graphing on the test?
ReplyDeleteyes, that is perfect.
ReplyDeleteSo about part b, the energy generated in 10 ms...is the "roughly" part of this equation meant to tell us that we don't have to integrate? Because I understand that P=RI^2, but I is decreasing as a function of time because the capacitor is losing its stored energy. So is the 10% accuracy just telling us to ignore the fact that I is decreasing with time?
ReplyDeletedo you mind posting the answer to part b of the first part about heat energy?
ReplyDeleteI'm so confused! In office hours yesterday it was said that Va-Vb= 0v because there is no current. But in class we did a similar problem (except gate between resistor and capacitor) and there was voltage and current at t=0. Was it that in the class example at t=0 the gate was closed? Could someone explain what is going on in the circuit when the gate is open (resistor, capacitor (charged), and at the marked points)?
ReplyDeleteI used V = Vi*(e^(-t/RC)) to find V and then used U = C(V^2)/2 and my answer is exactly half of the energy I would get with U = R(I^2)*t.
ReplyDeleteI don't see anything wrong with either solution. why are the answers different?