Solve RC Circuit Problems: 2.00μF Capacitor, 1.3 kΩ Resistor

AI Thread Summary
A 2.00μF capacitor discharges through a 1.3 kΩ resistor, with an initial charge of 5.10μC. To find the current 9.00ms after connection, the voltage across the capacitor is calculated as V0 = Q/C, resulting in 2.55V. The maximum current in the resistor is determined using Io = V0/R, yielding 301mA. The remaining charge on the capacitor after 8.00ms can be calculated using the discharging equations, but the initial voltage is crucial for accurate results. Proper unit checks are emphasized throughout the calculations.
mad_monkey_j
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Homework Statement



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?

Homework Equations



Discharging capacitor
I = (Io-Vo)e^-t/RC
q = Qe^-t/RC

C= Q/(Vf-Vi)

I = V/R

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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What is the voltage across the capacitor with its initial charge?
 
It should be:

V=C/Q

V=2/5.1 = 392mV
 
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?
 
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?
 
mad_monkey_j said:
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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