Discussion with teacher exponential growth vs decay

From my PAG book, I wrote that for the voltage against time graph for charging a capacitor, the graph should be an exponential growth curve. You told me it should be an exponential decay curve, I'm confused about that because I am still pretty sure it's an exponential growth curve. Can you explain what you meant by this?

I asked that to my teacher, he replied:
Yes, exponential growth happens when the power that the number (here ‘e’) is being raised by is greater than one, and decay when its less than one. Exponential growth will end up simply getting bigger and bigger in magnitude.

I'm still confused, so for charging a capacitor for the voltage against time*graph would you describe it as exponential growth or decay?
What does he want me to refer to it as, even though I still think it's exponential growth. Can someone help me understand what I should write it as please? Thank you!

Reply 1

Nitrotoluene

Original post

by LisaPerera

From my PAG book, I wrote that for the voltage against time graph for charging a capacitor, the graph should be an exponential growth curve. You told me it should be an exponential decay curve, I'm confused about that because I am still pretty sure it's an exponential growth curve. Can you explain what you meant by this?
I asked that to my teacher, he replied:
Yes, exponential growth happens when the power that the number (here ‘e’) is being raised by is greater than one, and decay when its less than one. Exponential growth will end up simply getting bigger and bigger in magnitude.
I'm still confused, so for charging a capacitor for the voltage against time*graph would you describe it as exponential growth or decay?
What does he want me to refer to it as, even though I still think it's exponential growth. Can someone help me understand what I should write it as please? Thank you!

As the capacitor is charging, the voltage increases to a value of V₀.
The voltage is given by the function:
V(t) = V₀(1 - e^(-t/τ))
where τ = RC.
The value of e^(-t/τ) decreases with time, but because it is subtracted from 1, the voltage increases.
An easy way to visualise this is: (1 - e^(-t/τ)) begins at 0 and increases to 1. At the start, it increases rapidly, then slows down as it approaches the end point.
What about the discharge? During the discharge, the voltage decreases: V (t) = V₀ e^(-t/τ), from V₀ to 0.
Briefly, the voltage goes up exponentially towards V₀ with a time constant RC. The important thing to note here is that the exponential e^(−t/τ) is the part that decreases.

Ciao,
Sandro

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