Capacitor Network: C1,C2 & C3 in Series & Parallel

Click For Summary
Capacitors C1 and C2 are connected in series, while this combination is in parallel with capacitor C3. To find the equivalent capacitance, the formula for capacitors in series (1/Ceq = 1/C1 + 1/C2) is applied first to C1 and C2, yielding a value that is then added to C3's capacitance (Ceq = x + C3). The problem requires calculating the net capacitance and the voltage across each capacitor when a voltage of 26.0 V is applied to the network. The specific capacitance values are C1 = 2.98 μF, C2 = 4.15 μF, and C3 = 1.64 μF. The solution involves determining both the equivalent capacitance and the voltage distribution across the capacitors.
GoldWing
Messages
8
Reaction score
0

Homework Statement


Capacitors C1 and C2 are connected in series and this combination is connected in parallel with capacitor C3.
HW8_2.jpg

Homework Equations


Q=CV
Ceq=C1+C2+C3 equivalent capacitance for capacitors in parallel
1/Ceq=1/C1+1/C2+1/C3 equivalent capacitance for capacitors in series

The Attempt at a Solution


The question says that C1 and C2 are connected in series, so I would use the second equation, so I would have 1/C=1/C1+1/C2, but then I don't know how I would add that to C3.
 
Physics news on Phys.org
Say you get x for the series connection and y is the capacitance of C3, the total capacitance equals x+y.
 
what is the problem statement?
 
rude man said:
what is the problem statement?
I'm sorry, I thought I had stated this. The question is: What is the net capacitance?
If 26.0 V is applied across the whole network, calculate the voltage across each capacitor. Use the numerical values of capacitance given, C1 = 2.98 μF, C2 = 4.15 μF, and C3 = 1.64 μF.
 
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
5
Views
646
  • · Replies 2 ·
Replies
2
Views
14K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 15 ·
Replies
15
Views
885
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K