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Assume a circuit like that in the picture is used to charge a 470-uF capacitor to 50...

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t4trendesetter | Honors

Posted August 22, 2013 at 10:11 AM via web

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Assume a circuit like that in the picture is used to charge a 470-uF capacitor to 50 volts, which is then discharged through a 1.8-k ohhm resistor. How do I find the time constant. I know it has something to do with T=RC...

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 9, 2013 at 3:32 PM (Answer #1)

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Let us suppose the capacitor C =470 microF has already been charged to a voltage U0= 50 Volts and is now connected in series with the resistor R =1.8 kOhms.

The voltage on the capacitor VC (equal to the voltage on the resistor VR) will be a function of time and will decrease with time as the capacitor is discharging.

We write now these two voltages (on the capacitor and on the resistor):

`V_C(t) = -(Q(t))/C = -(int(I*dt))/C`    and `V_R = I*R`

where the sign minus in the expression of VC comes from the fact that the capacitor is discharging with time.

and impose the condition that `V_C = V_R`

`I*R + (int(I*dt))/C =0`   

by differentiating and rearranging we obtain

`(dI)/dt +I/(R*C) =0` where `I =I(t)` is a function of time.

to solve we separate variables I and t

`(dI)/I = -dt/(RC)`

and by integration from t=0 to t (from I0 to I(t))

`ln((I(t))/I_0) = -t/(RC)`  or equivalent `I(t) =I_0*e^(-t/(RC)) =I_0*e^(-t/tau)`

We can now define the time constant `tau = R*C =1800*470*10^-6 =0.846 seconds`

as the time that the current needs to decrease to a value of (1/e) from its initial value I0.

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