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!
Well, two ways to do it, one with voltage, the other with charge. I would suggest the voltage approach since you already have the supply voltage given. Find out what proportion (percent wise) of the supply voltage is dropped across the capacitor when t = RC (use the same approach you used in part a) except now you know the time but you're looking for the percentage). Let's say the answer is p%. Figure out p% of 200V to get the potential difference across the capacitor at that time, and put that into the formula E = 1/2*C*V^2. Use the formula for power dissipation across a resistor which is P = V^2/R. You know the voltage drop across the capacitor at t = RC. You know the supply voltage. Since the capacitor and the resistor are in series, what can you say about the voltage across the resistor at this time? Hence work out the power loss using the formula.
Find the voltage to answer b using q=CV. That will help since you have capacitance and voltage value.
I didn't see.. :rofl: I thought it would help to read the stuff there. In fact for those interested, why not search up PF for all similar RC/capacitor problems and the discussions following them for info.
I just stumbled upon this post and have focused on this quote. In order to solve for the energy stored in the capacitor, I think of how the energy is stored--in the E field! 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'