
#1
Mar2911, 11:18 PM

P: 33

1. The problem statement, all variables and given/known data
A 2.00μF capacitor with an initial charge of 5.10μC is discharged through a 1.3 kΩ resistor. a) calculate the current through the resistor 9.00ms after the resistor is connected across the capacitor b) what charge remains on the capacitor after 8.00ms c) what is the maximum current in the resistor? 2. Relevant equations Discharging capacitor I = (IoVo)e^t/RC q = Qe^t/RC C= Q/(VfVi) I = V/R 3. The attempt at a solution C= Q/(VfVi) V= 5.1/2 V=2.55V I=V/R I=2.55/1300 I=1.9mA Unable to find Io and have no idea where to go from there. 



#2
Mar2911, 11:25 PM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,416

What is the voltage across the capacitor with its initial charge?




#3
Mar2911, 11:34 PM

P: 33

It should be:
V=C/Q V=2/5.1 = 392mV 



#4
Mar2911, 11:36 PM

P: 241

RC circuit problems
If you want to find I0, you should know V0. But what is V0?
Do you know the capacitor relation C=Q/V. Can you use this to find V0? So knowing V0 and R you can find I0. And can you check your first equation? how the term V0 exist there? 



#5
Mar2911, 11:56 PM

P: 33

So,
Io = Vo/R Io = 301mA I=(Io)e^t/RC (Io)e^t(650000000) (Io)e^1.38461538 × 1011 1 * Io I = 301mA? 



#6
Mar3011, 08:08 AM

Mentor
P: 11,441

Capacitance has units [Coul]/[Volt]. Charge has units [Coul]. So the basic expression is C = Q/V giving you V = Q/C . 


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