Find the charge of Capacitor 2 in the circuit

In summary: Since you've correctly said that the current through the capacitors is zero, just remove them from the circuit and keep the respective terminals open. How does the current flow in this new circuit?The current flows through the battery and the resistor. What's the potential difference across a resistor with zero current flowing?0 FOR THE RESISTOR since i is 0 but the battery itself still has a voltage? right?0 FOR THE RESISTOR since i is 0 but the battery itself still has a voltage? right?
  • #1
Tomi Kolawole
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0

Homework Statement


media%2F208%2F208d83af-d0cc-41dd-9adf-fd2111268b66%2FphpXJOsJE.png


Homework Equations


I am to use korchkoffs method to solve this problem

The Attempt at a Solution


I already solved for the overall current which is the same throughout because C1 and C2 impeded current flow in sections of the circuit and i have found the charge on c1 BUT I HAVE ISSues evluating the bottom right mini-circuit containing C2 to find its charge Q2.I don't know how write its korchoff equation
 

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  • #2
What potential differences do you already know in that subcircuit?
 
  • #3
all the Resistances Voltages and Capacitance are known
 
  • #4
Tomi Kolawole said:
all the Resistances Voltages and Capacitance are known
And what are they? Can you annotate a drawing of the subcircuit with the known values of potential differences?
 
  • #5
The values of the known circuit components are at the very bottom of the image.its easy to miss aha.

This is my Circuit diagram with the current direction and teminals laballed
uy9xXGv.png
 

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  • #6
Can you pencil in the known potential difference values on that diagram?
 
  • #7
ryPpiy4.png
V1 is not in this subcircuit that i made but you can see it in the overall diagram in the original image
 

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  • #8
Okay, let me clarify my request. What are the potential differences across the resistors in the subcircuit?
 
  • #9
Its V=IR so V=4/3 * 3 so the resistors have a V of 4 Volts
My Calculated I(current)= 4/3 and R is a given value of 3.
The capacitor is fully charged since the current (as stated in the question) has been flowing for awhile.SO i'M BASically wondering if the v2 AND R on the same branch as the capacitor (C2) will still contribute voltage even if there is no current flowing through their branch cause of the Fully charged Capacitor C2
 
Last edited:
  • #10
Tomi Kolawole said:
Its V=IR so V=4/3 * 3 so the resistors have a V of 4 Volts
My Calculated I(current)= 4/3 and R is a given value of 3.
Okay.
The capacitor is fully charged since the current (as stated in the question) has been flowing for awhile.SO i'M BASically wondering if the v2 AND R on the same branch as the capacitor (C2) will still contribute voltage even if there is no current flowing through their branch cause of the Fully charged Capacitor C2
Sure, voltage source always has a fixed potential difference. What's the potential difference across a resistor with zero current flowing?
 
  • #11
0 FOR THE RESISTOR since i is 0 but the battery itself still has a voltage? right?
 
  • #12
Tomi Kolawole said:
0 FOR THE RESISTOR since i is 0 but the battery itself still has a voltage? right?
Right. So you should be able to write KVL for the loop with the only unknown potential difference being that of the capacitor.

upload_2018-3-20_11-33-20.png
 

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  • #13
Tomi Kolawole said:
0 FOR THE RESISTOR since i is 0 but the battery itself still has a voltage? right?
Yes.

Since you've correctly said that the current through the capacitors is zero, just remove them from the circuit and keep the respective terminals open. How does the current flow in this new circuit?
 
  • #14
thank you so much kind sir!
 

1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This electric field stores energy in the form of electric charge.

2. How do you find the charge of a capacitor?

The charge of a capacitor can be found by multiplying the capacitance (C) of the capacitor by the voltage (V) applied to it. This can be represented by the equation Q = CV. The unit of charge is coulomb (C).

3. What is the role of a capacitor in a circuit?

Capacitors have several uses in a circuit. They are commonly used to store energy, filter out unwanted signals, and block direct current while allowing alternating current to pass through. They can also be used to regulate voltage, create time delays, and act as coupling devices between different parts of a circuit.

4. How do you calculate the charge of Capacitor 2 in a circuit?

To calculate the charge of Capacitor 2 in a circuit, you would first need to know the capacitance and voltage of the capacitor. Once you have this information, you can use the formula Q = CV to find the charge. Make sure to use the correct units for capacitance (farads) and voltage (volts) in the calculation.

5. What factors can affect the charge of a capacitor in a circuit?

The charge of a capacitor can be affected by several factors such as the capacitance of the capacitor, the voltage applied to it, the type of dielectric used, and the temperature. Additionally, the charge of a capacitor can also be affected by the resistance of the circuit and any other components connected to it.

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