# Homework Help: Voltage across capacitor with function given

1. Mar 10, 2014

### Maylis

1. The problem statement, all variables and given/known data
Problem statement in image

2. Relevant equations

3. The attempt at a solution
For this problem, I believe I am doing the math right, and getting infinite voltage at t=∞

However, in general there are so troubling facts to me in the problem statement. It is not so clear on the voltage boundaries. Also, I am not sure about these instantaneous voltage and current changes. I am so confused about my professor saying no instantaneous voltage change can occur. Okay, fine I know I(t) is proportional to dv/dt, and if dv/dt is infinite, then I(t) is infinite, which is not possible. However, how can you have anything other than an instantaneous change? I mean, how else does something change on a small enough scale other than instantaneously, isn't that the entire point of infinitesimal calculus?? What's it going to do, change gradually? How is that expressed?

I would actually rather say yes to both parts (a) and (b), because obviously when the switch turned on, stuff instantly began happening, which means both of those changed instantly, but whats holding me back is this whole business about no instantaneous voltage changes being allowed.

Does it mean at t < 0, V(t) = 20V, or is it t ≤ 0, V(t) = 20V

then t > 0 is given by the function in the problem statement.

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Last edited: Mar 10, 2014
2. Mar 10, 2014

### Staff: Mentor

The problem statement says that "The voltage across a 0.2 mF capacitor was 20 V until a switch in the circuit was opened". So the assumption is that for t < 0 the voltage is taken to be a constant 20 V. That's the "big picture" view of the situation.

Now, if you take a very close look at the time around the switching event then you can say that at time t = 0- the potential on the capacitor is 20 V, the switching event occurs at time t = 0, and that at time t = 0+ the potential across the capacitor is *still* 20 V because
$$\lim_{t \to 0^+}(60 - 40 e^{5t}) = 20 = \lim_{t \to 0^-}(60 - 40 e^{5t})$$
The function that represents the voltage on the capacitor is continuous around the switching event.

In practical terms it means that changing the voltage on a capacitor requires the physical movement of charges onto or off of the capacitor, which takes time to happen. The only way to make an instantaneous (no time expired) change of voltage across a capacitor would be if an infinite current were applied. There's no such beast in the real world with real components.

Now, if you had a discontinuous function where the limits from the left did not equal the limit from the right at a given instant, then you would have an instantaneous change. This in fact occurs for the function representing the current through the capacitor, which goes from zero to some finite value determined by the circuit components (think: time constant). The situation is the opposite for inductors, where the current function is continuous but the voltage function can make an instantaneous change: the inductor creates an emf that opposes a change in current.

Regarding your solution for part (d), Can you explain why you're integrating the voltage function? Aren't you just looking for the voltage at different times?

3. Mar 10, 2014

### Maylis

I see, for some reason I was thinking in my head that was the function for current, my head was in the clouds. I understand your explanation crystal clear, the limit from the left is the limit of the right, so its a continuous function. I was very confused by the explanation my professor gave me about the voltage not being allowed to be an instantaneous change, that cleared it up nicely for me.

I got 0.36 J for (d)

4. Mar 11, 2014

Looks good!