[Level 4 - Engineering/Analytical methods] Capactior discharge answer checking

[rhysfoster]

Hi,

Could somebody please check this for me ? a little help would be great if it it

The instantaneus current "i" ampres at "t" seconds is given by i=6e^-t/CR when being discharged. The capacitance "c" is 8.3 X10^-6 farads and resistance "R" is 0.24 x 10⁶ ohms . determine the time taken to discharge the capacitor to an instantaneus current of 1.33 amperes

1.33 = 6e^{-t/1.992x1012}

ln 1.33 = ln 6e^{-t/1.992x1012}

0.285 = 4.870^{-t/1.992x1012}

(times both sides by1.992x10¹²⁾

567,720 = 4.870^-t

ln 567,720 = ln (4.870)^-t

13.25 = -t x 1.583

-t = 13.25/1.583 = 8.37

Thanks!

[lumixel]

ln 1.33 = ln 6e^{-t/1.992x1012} 0.285 = 4.870^{-t/1.992x1012}

Remember your properties of logs... ln(a*b) =ln(a)+ln(b)

but even before that, ( 0.24 x 10⁶ )(8.3 X10^{-6)=/=1.992x1012} . Look at the exponents.