Solve RC Circuit Problem: Calculate Charge on Capacitor

In summary: Well then it would be -RC*ln(1/4) = -(50)(2)*ln(1/4)= 138.63 ms but that is somehow wrong.. :/Oh stupid me! I freakin' forgot to convert the units... it should be 138.63 microseconds or .1386 milliseconds. Thanks for the help! :P
  • #1
hover
343
0

Homework Statement


What is the charge on the capacitor?


Homework Equations


Kirchoff's rule
change in V =q/c


The Attempt at a Solution


I want to use Kirchoff rule but I don't know if this is the right way to go.
 

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  • #2
When the capacitor is completely charges, there is no current in that branch. The voltage across the capacitor is the potential difference across 40 ohm resistor.
 
  • #3
rl.bhat said:
When the capacitor is completely charges, there is no current in that branch. The voltage across the capacitor is the potential difference across 40 ohm resistor.

Ya, I was able to figure out that the charge on the resistor would be 80 microCoulombs . The voltage difference between the plates of the capacitor was 40 volts and from there it was plugging in. Now I face another problem where I can't find how long it will take the capacitor to discharge to 25% of 80 microCoulombs. I derived that it should be -RC*ln(1/4)=t. I plug in R= 50, c = 2 and get the wrong answer :/
 
  • #4
hover said:
Ya, I was able to figure out that the charge on the resistor would be 80 microCoulombs . The voltage difference between the plates of the capacitor was 40 volts and from there it was plugging in. Now I face another problem where I can't find how long it will take the capacitor to discharge to 25% of 80 microCoulombs. I derived that it should be -RC*ln(1/4)=t. I plug in R= 50, c = 2 and get the wrong answer :/
t = 2 μF.
 
  • #5
rl.bhat said:
t = 2 μF.

How does t = 2 MICROFARADS? Don't you mean seconds or milliseconds? I don't quite understand your logic though. Doesn't the capacitor discharge through 2 resistors that become 50 ohms?
 
  • #6
hover said:
How does t = 2 MICROFARADS? Don't you mean seconds or milliseconds? I don't quite understand your logic though. Doesn't the capacitor discharge through 2 resistors that become 50 ohms?
Sorry.It is typo. I mean C = 2 μF. Your R is correct. You didn't mention your answer.
 
  • #7
rl.bhat said:
Sorry.It is typo. I mean C = 2 μF. Your R is correct. You didn't mention your answer.

Well then it would be -RC*ln(1/4) = -(50)(2)*ln(1/4)= 138.63 ms but that is somehow wrong.. :/
 
  • #8
Oh stupid me! I freakin' forgot to convert the units... it should be 138.63 microseconds or .1386 milliseconds.

Thanks for the help! :P
 

FAQ: Solve RC Circuit Problem: Calculate Charge on Capacitor

How do you calculate the charge on a capacitor in an RC circuit?

The charge on a capacitor in an RC circuit can be calculated using the formula Q = Q0(1-e-t/RC), where Q0 is the initial charge on the capacitor, t is the time, R is the resistance, and C is the capacitance.

What is an RC circuit and how does it work?

An RC circuit is a type of electrical circuit that contains a resistor (R) and a capacitor (C). It works by allowing the capacitor to charge through the resistor, creating a time-varying voltage across the capacitor. This voltage can then be used to power other components in the circuit.

What is the significance of the resistance and capacitance values in an RC circuit?

The resistance (R) and capacitance (C) values in an RC circuit determine the rate at which the capacitor charges and discharges. A higher resistance will result in a slower charging/discharging rate, while a higher capacitance will result in a longer time for the capacitor to reach its maximum charge.

How does the charge on the capacitor change over time in an RC circuit?

The charge on the capacitor in an RC circuit follows an exponential curve, starting from an initial charge (Q0) and approaching a maximum charge (Qmax) as time goes on. The rate of change of charge on the capacitor is dependent on the values of R and C in the circuit.

Can you solve an RC circuit problem without knowing the initial charge on the capacitor?

Yes, it is possible to solve an RC circuit problem without knowing the initial charge on the capacitor. This can be done by using the equation Q = Qmax(1-e-t/RC). In this case, the initial charge (Q0) is assumed to be 0.

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