Voltage across capacitor with function given

In summary, the problem statement is confusing and does not clearly state the boundaries for the voltage and current changes. It is not clear if t ≤ 0 is when the switch is opened or when the potential across the capacitor is 20 V. The time around the switching event is continuous, so the voltage on the capacitor is 20 V at that time.
  • #1
gfd43tg
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Homework Statement


Problem statement in image

Homework Equations


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 what's 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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  • #2
Maylis said:

Homework Statement


Problem statement in image

Homework Equations


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 what's 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.

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?
 
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  • #3
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
Maylis said:
I got 0.36 J for (d)
Looks good!
 
  • #5


I understand your confusion and concerns about the problem statement. It is important to note that in real-world situations, instantaneous changes are not physically possible. However, in theoretical calculations, we often assume instantaneous changes for simplicity. In this case, the function given for voltage across the capacitor may suggest an instantaneous change, but it is important to remember that this is just a theoretical calculation and in reality, the change would occur gradually.

Regarding the voltage boundaries, it is not explicitly stated in the problem statement, but we can assume that the voltage at t=0 is 20V, and at t=∞ it would approach infinity. This is because as the capacitor charges, the voltage across it increases, and in theory, it would never reach a steady state.

In conclusion, while it may seem counterintuitive, the concept of no instantaneous voltage changes is a fundamental principle in electrical engineering and is used to simplify theoretical calculations. It is important to keep this in mind while solving problems and to remember that in real-world situations, changes occur gradually.
 

FAQ: Voltage across capacitor with function given

What is the function of a capacitor?

A capacitor is an electronic component that is used to store electrical charge. It is made up of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied, one plate accumulates a positive charge while the other accumulates a negative charge, creating an electric field between them.

How does the voltage across a capacitor change over time?

The voltage across a capacitor increases as it charges and decreases as it discharges. The rate at which the voltage changes is determined by the capacitance of the capacitor and the current flowing through it.

What is the relationship between the voltage and current across a capacitor?

The voltage and current across a capacitor are inversely proportional. This means that as the voltage increases, the current decreases, and vice versa. This relationship is described by the equation V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance.

How do you calculate the voltage across a capacitor with a given function?

To calculate the voltage across a capacitor with a given function, you can use the equation V = V0(1-e^(-t/RC)), where V0 is the initial voltage, t is time, R is the resistance, and C is the capacitance. This equation is derived from the charging equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

What factors affect the voltage across a capacitor?

The voltage across a capacitor is affected by the capacitance, the resistance, and the time. A larger capacitance will result in a higher voltage, while a larger resistance will result in a lower voltage. Additionally, the voltage will change over time as the capacitor charges and discharges.

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