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

In summary: You can verify your answer by using a calculator.In summary, the conversation was about a 2.00μF capacitor with an initial charge of 5.10μC being discharged through a 1.3 kΩ resistor. The questions asked were to calculate the current through the resistor 9.00ms after the resistor is connected across the capacitor, what charge remains on the capacitor after 8.00ms, and what is the maximum current in the resistor. The equations used were for discharging a capacitor, calculating capacitance, and for finding current through a resistor. The solution involved finding the voltage across the capacitor, using the capacitor relation C=Q/V to find the initial voltage, and then using that to
  • #1
mad_monkey_j
33
0

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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  • #2
What is the voltage across the capacitor with its initial charge?
 
  • #3
It should be:

V=C/Q

V=2/5.1 = 392mV
 
  • #4
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
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?
 
  • #6
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 .
 

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

How do I calculate the total resistance in this RC circuit?

To calculate the total resistance in an RC circuit, you need to add the resistance of the resistor (R) and the reactance of the capacitor (Xc). In this case, the resistor has a value of 1.3 kΩ and the capacitor has a value of 2.00μF. The formula for reactance is Xc = 1/(2πfC), where f is the frequency in Hertz and C is the capacitance in Farads. Since the frequency is not given in this problem, we can assume it is in a DC circuit and therefore the reactance is 0. To calculate the total resistance, we simply add 1.3 kΩ and 0, giving us a total resistance of 1.3 kΩ.

What is the time constant of this RC circuit?

The time constant of an RC circuit is the time it takes for the capacitor to charge to 63.2% of its maximum charge. It is calculated by multiplying the resistance (R) and the capacitance (C) in the circuit. In this case, the time constant would be 1.3 kΩ * 2.00μF = 2.6 milliseconds.

How do I calculate the charge on the capacitor after a certain amount of time?

The charge on the capacitor can be calculated using the formula Q = Qmax * (1 - e^(-t/RC)), where Qmax is the maximum charge, t is the time in seconds, R is the resistance in Ohms, and C is the capacitance in Farads. In this problem, Qmax can be calculated by multiplying the voltage (V) by the capacitance (C). So if we wanted to find the charge on the capacitor after 5 milliseconds, the formula would be Q = (2.00μF * 5V) * (1 - e^(-5ms/2.6ms)) = 9.71 μC.

How do I calculate the voltage across the capacitor at a specific time?

The voltage across the capacitor can be calculated using the formula Vc = Vmax * (1 - e^(-t/RC)), where Vmax is the maximum voltage, t is the time in seconds, R is the resistance in Ohms, and C is the capacitance in Farads. In this problem, Vmax can be calculated by multiplying the voltage (V) by the charge on the capacitor (Q). So if we wanted to find the voltage across the capacitor after 3 milliseconds, the formula would be Vc = (5V * 9.71 μC) * (1 - e^(-3ms/2.6ms)) = 4.46V.

How do I calculate the current in this RC circuit?

The current in an RC circuit can be calculated using Ohm's Law (I = V/R) or using the formula I = I0 * e^(-t/RC), where I0 is the initial current and t is the time in seconds. In this problem, the initial current can be calculated by dividing the initial voltage (V) by the total resistance (R). So if we wanted to find the current after 2 milliseconds, the formula would be I = (5V/1.3 kΩ) * e^(-2ms/2.6ms) = 3.58 mA.

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