Voltage across Capacitor equation derivation. Pls help, much appreciated

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nhtw
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Size:  38.8 KBso i came across this on I want to study engineering. I have drawn arrows at the point where they times both sides by -1 so that a decay exponential ie e^-1/RC is formed. My question is, mathematically what makes you do this. It could turn into an exponential increase if you don't right? so is it done out of intuition or math?? THANKS
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16Characters....
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You will get the exact same formula even if you do not multiply by -1. I think it might be because it makes the LHS integral a tiny bit nicer (though it really makes no difference in my opinion):

\displaystyle \int_0^{V_c} \frac{dV_c}{V - V_c} = \displaystyle \int_0^t \frac{dt}{RC}
 - \ln |V - V_c| + \ln |V| = \frac{t}{RC}
 \ln |V - V_c| = \ln |V| - \frac{t}{RC}
 V - V_c = Ve^{\frac{-t}{RC}} \implies V_c = V \left(1- e^{\frac{-t}{RC}} \right)

Now is probably the time where a physicist comes in an tells me it is not as simple as the maths suggests haha.
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uberteknik
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(Original post by 16Characters....)
You will get the exact same formula even if you do not multiply by -1. I think it might be because it makes the LHS integral a tiny bit nicer (though it really makes no difference in my opinion):

\displaystyle \int_0^{V_c} \frac{dV_c}{V - V_c} = \displaystyle \int_0^t \frac{dt}{RC}
 - \ln |V - V_c| + \ln |V| = \frac{t}{RC}
 \ln |V - V_c| = \ln |V| - \frac{t}{RC}
 V - V_c = Ve^{\frac{-t}{RC}} \implies V_c = V \left(1- e^{\frac{-t}{RC}} \right)

Now is probably the time where a physicist comes in an tells me it is not as simple as the maths suggests haha.
Nah, you got it in one.

The equation tells us the charge stored in an initially isolated capacitor (Vc = Vo) will exponentially decay at a rate governed by the product between the capacitance and current path resistance.
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