Is this integration remotely correct?

Scott92

Scott92

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Jun 11, 2022
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Picture1.png Im not sure its correct from the -EC equation above where says am i being paranoid or am i useless any help welcomed thanks
 
your image is a bit small and fuzzy for these old eyes, but a mistake I noticed is your calculation of the time constant, RC.

[imath]10 \, k\Omega = 10 \times 10^3 \, \Omega \text{ not } 10 \cdot 10^4[/imath]

[imath]RC = (10^4 \Omega) \cdot (22 \times 10^{-6} \, f) = 22 \times 10^{-2}[/imath]

If I didn’t err in substitution of the given values, I calculate the charge at [imath]t=0.5[/imath] as [imath]1.58 \times 10^{-4}[/imath] coulombs.
 
skeeter said:
your image is a bit small and fuzzy for these old eyes, but a mistake I noticed is your calculation of the time constant, RC.

[imath]10 \, k\Omega = 10 \times 10^3 \, \Omega \text{ not } 10 \cdot 10^4[/imath]

[imath]RC = (10^4 \Omega) \cdot (22 \times 10^{-6} \, f) = 22 \times 10^{-2}[/imath]

If I didn’t err in substitution of the given values, I calculate the charge at [imath]t=0.5[/imath] as [imath]1.58 \times 10^{-4}[/imath] coulombs.
Click to expand...
Argh yeah it wouldn't recognise most of the symbols so thought take a pic. Thanks for the help :thumbup::thumbup:
 
skeeter said:
your image is a bit small and fuzzy for these old eyes, but a mistake I noticed is your calculation of the time constant, RC.

[imath]10 \, k\Omega = 10 \times 10^3 \, \Omega \text{ not } 10 \cdot 10^4[/imath]

[imath]RC = (10^4 \Omega) \cdot (22 \times 10^{-6} \, f) = 22 \times 10^{-2}[/imath]

If I didn’t err in substitution of the given values, I calculate the charge at [imath]t=0.5[/imath] as [imath]1.58 \times 10^{-4}[/imath] coulombs.
Click to expand...
Sorry how do u get the 1.58 unsure what the err means ?
 
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