The discussion revolves around charging a 470µF capacitor to 330V using a 3V battery at a current of 10µA. Participants are exploring the time required for charging, the stored energy in the capacitor, and the current delivered by the battery.
There is an ongoing exploration of the problem with participants providing calculations and questioning their accuracy. Some guidance has been offered regarding the need to use the correct values and to confirm the current specified in the problem. Multiple interpretations of the voltage to use in calculations are being discussed.
Participants note the potential for errors due to exhaustion and the need for clarity on the capacitor's value and the voltage to be used in calculations. The original problem statement specifies the capacitor's charge voltage as 330V, which is a point of confusion for some.
elips said:Well, Q = CV
Q= 10*10^-6*3V
Q=30µC
i = dq/dt
t= 30*10^-6/10*10^-6 Is it true?
Wc= 0.5*C*V^2
Wc= 0.5*10*10^-6*(3^2)
Wc= 4.5 × 10-5µJ Is it true?
I'm not really sure about my answers...
gneill said:I thought that the capacitor was 470µF?
Well, it's true that a the time required to move a charge Q with a constant current I is
t = Q/I
You'll want to straighten out your capacitor value in your calculation.
gneill said:I thought that the capacitor was 470µF?
Well, it's true that a the time required to move a charge Q with a constant current I is
t = Q/I
You'll want to straighten out your capacitor value in your calculation.
gneill said:Your methodology is okay, you just need to use the correct values.
Capacitor: 470µF charged to 330V at a rate of 10µA
(That's a pretty small current; it'll take a looong time to charge! You might want to confirm this value with the original problem text).