# Discharging capacitor half life

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

$$v = V_0 \, e^{-\frac{t}{RC}}$$

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

$$v = V_0\, 2^{-\frac{t}{\log(2) \, RC}}$$

where "log" is the natural logarithm.

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

Last edited:
Thanks Uart,

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

$$v = V_0 \, e^{-\frac{t}{RC}}$$

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

$$v = V_0\, 2^{-\frac{t}{\log(2) \, RC}}$$

where "log" is the natural logarithm.

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

uart
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.

Delta2
Homework Helper
Gold Member
Half life of a quantity is the time it needs so that the quantity is reduced to half of its original value.

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 $$V_0$$ to $$\frac{V_0}{2}$$

Thanks.

sophiecentaur
you can see that the "half life" is $RC/\log_e(2)$
Just correcting a typo above. That should of course have been $RC \, \log_e(2)$