RC circuit problems


by mad_monkey_j
Tags: capacitors, electronics, rc circuit
mad_monkey_j
mad_monkey_j is offline
#1
Mar29-11, 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 = (Io-Vo)e^-t/RC
q = Qe^-t/RC

C= Q/(Vf-Vi)

I = V/R
3. The attempt at a solution

C= Q/(Vf-Vi)
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.
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SammyS
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#2
Mar29-11, 11:25 PM
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What is the voltage across the capacitor with its initial charge?
mad_monkey_j
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#3
Mar29-11, 11:34 PM
P: 33
It should be:

V=C/Q

V=2/5.1 = 392mV

n.karthick
n.karthick is offline
#4
Mar29-11, 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?
mad_monkey_j
mad_monkey_j is offline
#5
Mar29-11, 11:56 PM
P: 33
So,

Io = Vo/R
Io = 301mA
I=(Io)e^-t/RC

(Io)e^-t(650000000)
(Io)e^-1.38461538 10-11
1 * Io
I = 301mA?
gneill
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#6
Mar30-11, 08:08 AM
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P: 11,407
Quote Quote by mad_monkey_j View Post
It should be:

V=C/Q

V=2/5.1 = 392mV
Check your units. Always check your units!

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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