contestada

The capacitor in an circuit is discharged with a time constant of 40.4 s. At what time, in microseconds, after the discharge begins is the charge on the capacitor reduced to half its initial value?

Respuesta :

Apply:

[tex]Q=Q_0e^{-\frac{t}{RC}}[/tex]

where:

RC= time constant = 40.4 s

Q0 = initial charge

Q = final charge

[tex]\frac{Q_0}{2}=Q_0e^{-t/RC}[/tex][tex]\frac{Q_0}{2Q_0}=e^{-t/RC}[/tex][tex]\frac{1}{2}=e^{-t/RC}[/tex][tex]ln(\frac{1}{2})=-\frac{t}{RC}[/tex][tex]ln(\frac{1}{2})=-\frac{t}{40.4}[/tex][tex]-0.693\text{ = -}\frac{t}{40.4}[/tex][tex]t=\text{ 28 s}[/tex]

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