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Energy in a capacitor in an RC circuit. 
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#1
May2006, 06:01 AM

P: 230

The question asks:
For a charging circuit, the emf supplied by the battery is 200 V, R = [tex]2*10^5\omega[/tex] and C =[tex] 50 \mu F[/tex]. Find: a) the time taken for the charge to rise to 90% of the final value b) the energy stored in the capacitor at t=RC c) the power loss in R at t=RC I've worked out a). That was simple enough. However when it comes to b) and c) I'm having problems. For b) I want to use that [tex]E = \frac{CV^2}{2}[/tex], or one of its rearrangements (most probably the [tex]\frac{Q^2}{2C}[/tex] version), however whilst I can find a value for Q in terms of [tex]Q_{0}[/tex], this still leaves me with an unknown in the answer ([tex]Q_{0}[/tex]). And for c) I have no idea whatsoever  nowhere in my notes or text book, does it say anything about the energy or power in an RC circuit. Any hints (or blatent worked answers ) would be much appreciated! 


#2
May2006, 06:15 AM

HW Helper
P: 2,950




#3
May2006, 07:34 AM

P: 67

Find the voltage to answer b using q=CV. That will help since you have capacitance and voltage value.



#4
May2006, 10:53 AM

P: 1,780

Energy in a capacitor in an RC circuit.
You might also want to see http://www.physicsforums.com/showthr...783#post993783.



#5
May2006, 03:12 PM

P: 67

That's my post!



#6
May2006, 10:39 PM

P: 1,780




#7
Mar1810, 11:07 PM

P: 60

If E = 1/2 CV^{2}, then just plug in V(t) into that equation where t = RC. I hope that makes sense because I just banged it out on my homework. ...i'm just sayin' 


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