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## Main Question or Discussion Point

What is meant by discharging capacitor half life (the description). I seem to be getting different description, I would just like for someone to confirm it here for me please.

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What is meant by discharging capacitor half life (the description). I seem to be getting different description, I would just like for someone to confirm it here for me please.

- #2

uart

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If discharged through a resistor the capacitor voltage reduces exponentially via the equation

[tex] v = V_0 \, e^{-\frac{t}{RC}}[/tex]

Mathematically it's easy to represent an exponential of one base in other other base.

In this case the above exponential can be re-written as

[tex] v = V_0\, 2^{-\frac{t}{\log(2) \, RC}}[/tex]

where "log" is the natural logarithm.

From the above equation you can see that the "half life" is [itex]RC/\log_e(2)[/itex]

[tex] v = V_0 \, e^{-\frac{t}{RC}}[/tex]

Mathematically it's easy to represent an exponential of one base in other other base.

In this case the above exponential can be re-written as

[tex] v = V_0\, 2^{-\frac{t}{\log(2) \, RC}}[/tex]

where "log" is the natural logarithm.

From the above equation you can see that the "half life" is [itex]RC/\log_e(2)[/itex]

Last edited:

- #3

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I understand that, how would you describe half life (not mathematically or through equations).

If discharged through a resistor the capacitor voltage reduces exponentially via the equation

[tex] v = V_0 \, e^{-\frac{t}{RC}}[/tex]

Mathematically it's easy to represent an exponential of one base in other other base.

In this case the above exponential can be re-written as

[tex] v = V_0\, 2^{-\frac{t}{\log(2) \, RC}}[/tex]

where "log" is the natural logarithm.

From the above equation you can see that the "half life" is [itex]RC/\log_e(2)[/itex]

- #4

uart

Science Advisor

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Well obviously, it's the time that you have to wait until the voltage is half of it's original value. That's how I'd describe it.Thanks Uart,

I understand that, how would you describe half life (not mathematically or through equations).

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In the example of Uart, half life of the voltage is the time it gets for the voltage to reduce to the half of its starting value , that is the time it needs to go from [tex]V_0[/tex] to [tex] \frac{V_0}{2}[/tex]

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Thanks.

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sophiecentaur

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- #8

uart

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you can see that the "half life" is [itex]RC/\log_e(2)[/itex]

Just correcting a typo above. That should of course have been [itex]RC \, \log_e(2)[/itex]

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