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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]

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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]

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uart

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Thanks Uart,

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

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.

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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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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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