Finding charge of a capacitor in RC circuit

In summary, the charge on the 4 µF capacitor in this circuit would be 23.75 µC, calculated using the formula Q = CV = (4µ)(5.9375), with the voltage across the 4 Ω resistor being the same as the voltage across the capacitor and the current flowing through the resistor being the same as the current throughout the circuit.
  • #1
Teptip
4
0
Thanks, solved!
 
Last edited:
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  • #2
Teptip said:
Here is the circuit in question: http://imgur.com/glr2C4f

Homework Statement


Find the charge on the 4 µF capacitor if
R1 = 11 Ω and R2 = 4 Ω .
Answer in units of µC


Homework Equations


V = IR
C = Q/V


The Attempt at a Solution


We've never had to deal with capacitors in parallel with resistors before so I'm not sure how this works.

Do I assume the total resistance of the circuit is in series and thus = 16Ω ?

Yes, after a short time that the circuit was connected the capacitor is charged and does not take current any more.

Teptip said:
What voltage do I use for the C=Q/V equation?

The capacitor is connected parallel to the 4 ohm resistor. What do you know about the voltage across parallel connected circuit elements?

ehild
 
  • #3
Okay so I can calculate current I=V/R = 19/16 = 1.1875 A

V total = (V of Capacitor) + (V of 11Ω Resistor)

V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625

19 = (V of Capacitor) + 13.0625
V of Capacitor = 5.9375

Then plugging into the capacitance formula:
Q = CV = (4µ)(5.9375) = 23.75µ

Does this work look good?
 
  • #4
Teptip said:
Okay so I can calculate current I=V/R = 19/16 = 1.1875 A

The current is correct.

Teptip said:
V total = (V of Capacitor) + (V of 11Ω Resistor)
No, what do you mean of Vtotal? And you do not need it.

Teptip said:
V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625
You know the voltage across the 11 Ω resistor. Look at the figure. Where are the plates of the capacitor connected to?

ehild
 
  • #5
Is the voltage I want then just the voltage across the 4Ω resistor?
 
  • #6
Teptip said:
Is the voltage I want then just the voltage across the 4Ω resistor?

Yes. And what current flows though it?ehild
 
  • #7
Ah is it the same current as throughout, since the capacitor is quickly charged?

(I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)
 
  • #8
Teptip said:
Ah is it the same current as throughout, since the capacitor is quickly charged?

(I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)

Yes, they are connected to the same points, so the voltage must be the same across both. (The voltage does not "go through", it is across two points)

ehild
 

1. 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 = CV, where Q is the charge, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

2. What is the significance of the time constant in an RC circuit?

The time constant in an RC circuit is a measure of how quickly the capacitor charges or discharges. It is equal to the product of the resistance and capacitance in the circuit and determines the rate at which the capacitor reaches its maximum charge or discharges to zero.

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

The charge on a capacitor in an RC circuit follows an exponential curve, where it starts at zero and increases as the capacitor charges. As the capacitor reaches its maximum charge, the rate of increase slows down until it reaches a steady state.

4. Can the charge on a capacitor in an RC circuit be negative?

No, the charge on a capacitor in an RC circuit cannot be negative. The capacitor can only hold a positive charge, and any attempt to charge it with a negative voltage will result in the capacitor discharging.

5. What factors affect the charge on a capacitor in an RC circuit?

The charge on a capacitor in an RC circuit is affected by the capacitance of the capacitor, the voltage across the capacitor, and the resistance in the circuit. A higher capacitance or voltage will result in a larger charge, while a higher resistance will slow down the charging process.

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