Another one of them circuit questions

  • Thread starter Thread starter NotaPhysicsMan
  • Start date Start date
  • Tags Tags
    Circuit
Click For Summary
The discussion revolves around calculating the charge on a capacitor in a circuit with two resistors, a capacitor, and a battery. The user initially struggles to find the voltage drop across the components but eventually determines the current using the equation 12V = 2I + 4I, leading to a current of 2A. They calculate the voltage across the 4-ohm resistor as 8V. Using the formula q = CV, with the capacitance of 9.0μF, they find the charge on the capacitor to be 7.2 x 10^-5 C. The user expresses satisfaction with their resolution of the problem.
NotaPhysicsMan
Messages
146
Reaction score
0
The circuit in the drawing shows two resistors, a capacitor and a battery. When the capacitor is fully charged, what is the magnitude of q of the charge on one of its plates?

Here's what I got so far:

So 12= 2I + 4I

Since q=CV and I have C=9.0uF, all I need is the voltage. And since the voltage drop between parallel circuits is the same then all I have to find is the voltage at the node going into 4 ohm resistor? Sounds like a good method but I can't seem to get the voltage drop.

Any ideas? :rolleyes:
 

Attachments

  • Capac.JPG
    Capac.JPG
    3.4 KB · Views: 1,185
Physics news on Phys.org
okay for those who can't see the attachment because it's waiting approval:

Code:
|-R=2.0ohm-|--------|
|                |           |
V=12V        R=4ohm  C=9.0uF 
|                |           |
|__________|_______|

Ok where I placed the letters is where you'll get a resistor or capacitor or voltage etc.
ugh, the code thing doesn't work so well, but the R=4ohm is in the middle branch and the C=9.0uF is at the end branch.
 
Ack, I got it nvm.

Easy really, almost had it too

12=2I+4I, I=2

V=IR at the 4
V=2*4
V=8 Volts

q=CV
=9.0x10^-6Fx8Volts
=7.2x10^-5 C.

Thanks notaphysicsman :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
20
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K