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The following expression called transient in Electronics, which allows you to calculate several variables such time, voltage, capacity, etc.

This is a current time function without numbers, where A is a constant, t is a time variable and T as well.

$$i(t)=A\left(1-e^{-\frac t T}\right)$$

with numbers:

$$6.32=10\left(1-e^{-\frac {-10\cdot 10^{-3}} T}\right)$$

I'd like to convert it to natural logarithm in order to get T, and I'm not sure I did it well, but I write my solution:

$$\frac{-10\cdot 10^{-3}}{\ln\left(-1\left(-1+\frac{6.32}{10}\right)\right)}=T$$

Thanks for the solution and reply, I'd be grateful for it!

user
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Norbert Kts
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  • This is hard to read, proper formatting would help. here is a good tutorial on formatting for this site. Absent that, at least add some parentheses. Does e^-t/T mean $e^{-t/T}$ ? $\frac {e^{-t}}T$? Something else? – lulu Jan 22 '24 at 13:56
  • We can start with $i(t)=A\left(1-e^{-\frac t T}\right) \implies e^{-\frac t T}=1-\frac{i(t)}A$. – user Jan 22 '24 at 13:59

1 Answers1

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Checking your solution we have

$$i(t)=A\left(1-e^{-\frac t T}\right) \iff e^{-\frac t T}=1-\frac{i(t)}A$$

then tacking natural logarithm both sides we obtain

$$-\frac t T=\log\left(1-\frac{i(t)}A\right) \iff T=-\frac{t}{\log\left(1-\frac{i(t)}A\right)}$$

and with numbers

$$T=-\frac{-10\cdot 10^{-3}}{\log\left(1-\frac{6.32}{10}\right)}$$

which agrees with your solution.

user
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