Equivalent Capacitors and their charges:

AI Thread Summary
The discussion revolves around solving a capacitor circuit problem where the equivalent capacitance (C-eq) is given as 9.22 microfarads. The user initially attempted to combine capacitors in parallel and series but encountered negative values, indicating an error in their approach. It was clarified that the correct configuration involves the 12μF and 8.35μF capacitors in series, which are then in parallel with a 4.5μF capacitor. This combination is in series with an unknown capacitor C, and the entire setup is parallel to a 7.22μF capacitor. The user is advised to write the correct equations based on this configuration to solve for C.
spriter
Messages
3
Reaction score
0

Homework Statement


Hi, I'm trying to solve this problem on MasteringPhysics:

This circuit is given, and it says C-eq for the circuit is 9.22 microfarads:

http://img143.imageshack.us/img143/3475/walker4ech21pr118.jpg

The question is to find C in the figure knowing C-eq is 9.22 microfarads.

Homework Equations


C-eq series = 1/C-eq = 1/C1 + 1/C2 + ...
C-eq parallel = C1 + C2 + C3 ...

The Attempt at a Solution



At first I tried to take the three center capacitors to be parallel (The 7.22, 4.25, and 8.35) and find the Ceq of them and then take the other two to be in a series with this C-eq. But when I do that, I get a negative value for C.

Then I tried to take the C, 12.0 uF, and 8.35 uF to be in a series, combine them and then have that C-eq be parallel to the other two, but that also gives me a negative value.


Thank you.
 
Last edited by a moderator:
Physics news on Phys.org
In the circuit diagram no three capacitors are either series or parallel.
Here 12μF and 8.35μF are in series. Equivalent of these capacitors is in parallel with 4.5 μF.
Combination of these capacitors is in series with C. Finally all the above is in parallel with 7.22 μF. The whole combination is equal to 9.22 μF. Write down the correct equations and solve for C.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top