Respuesta :

skyluke89
The charge Q at time t on the capacitor (initially charged) in a RC-circuit is given by
[tex]Q(t)=Q_0 e^{-t/\tau}[/tex] (1)
where [tex]Q_0[/tex] is the initial charge, and [tex]\tau =RC[/tex] is the time constant of the circuit.

Using [tex]C=20 \mu F=20 \cdot 10^{-6}F[/tex] and [tex]R=18 k\Omega = 18 \cdot 10^3 \Omega[/tex] we can calculate the time constant:
[tex]\tau = RC= 0.36 s[/tex]

The problem asks for the time t at which the capacitor's charge is reduced to [tex]Q(t)=15 \mu C[/tex]. Since the initial charge is [tex]Q_0 = 30 \mu C[/tex], we can write
[tex] \frac{Q(t)}{Q_0}= \frac{15 \muC}{30 \muC} = \frac{1}{2} [/tex]
And we can rewrite equation (1) as
[tex] \frac{1}{2} = e^{-t/\tau} [/tex]

By solving this and by using [tex]\tau=0.36 s[/tex], we find
[tex]t=-\tau \ln( \frac{1}{2} )=0.25 s[/tex]
okpalawalter8

The time does it take to reduce the capacitor's charge to 15.0 μc is mathematically given as

t=0.25 s

What is time does it take to reduce the capacitor's charge to 15.0 μc?

Question Parameters:

A 20.0 μf capacitor initially charged to 30.0 μc is discharged through a 1.80  resistor.

Generally, the equation for the Charge   is mathematically given as

[tex]Q(t)=Q_0 e^{-t/\tau}[/tex]

Therefore

[tex]\frac{Q(t)}{Q_0}= \frac{15 \muC}{30 \muC}[/tex]

[tex]\frac{Q(t)}{Q_0}= \frac{1}{2}[/tex]

In conclusion, where \tau is

[tex]\tau=RC\\\\\tau=20.0 u*1.80 \\\\\tau=0.36[/tex]

Hence

[tex]t=-\tau \ln( \frac{1}{2} )[/tex]

t=0.25 s

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