I keep getting the first part wrong and I can\'t do the rest if I have that wron

Question

I keep getting the first part wrong and I can't do the rest if I have that wrong. HELP!

For the system of capacitors shown in the figure below, find the following. (Let C1 = 8.00 µF and C2 = 8.00 µF.)

(a) the equivalent capacitance of the system

Your response differs from the correct answer by more than 10%. Double check your calculations. µF

(b) the charge on each capacitor

(c) the potential difference across each capacitor

Explanation / Answer

1)

capacitors in series:

1/Ct=1/C1 + 1/C2 + 1/C3 ...

Capacitors in parallel:

Ct=C1 + C2 + C3 ...

The top portion is in series, so

1/Ct=1/8F + 1/6F=7/24=3.43F

Likewise, the bottom portion is in series:

1/Ct=1/2F + 1/8F=5/8=1.6F

These two equivalencies are now in parallel, so add them for the total equivalence:

3.43F + 1.6F=5.03F

2)

For capcitors in series, the charge is the same for each capcitor, and the the voltage of the individual capacitors must add to the total voltage.

Coversely, for capacitors in parallel, the charge of individual capacitors must equal the total charge, and the voltage is the same for each capacitor.

--the charge for the top equivalence (3.43F) and the bottom equivalence (1.6F) knowing that the voltage running through both=90V

equivalance 1:

Q=(3.43F)(90V)=307.8C

equivalence 2:

Q=(1.6F)(90V)=144C

Now you can expand the top and bottom equivalence and find the charge, which will be the same as what you just found (because they're in series):

C1--> Q=307.8 C

6.00F--> Q=307.8C

2.00F--> Q=144C

C2 --> Q=144F

3)

For potential difference, use the charges just found:

V=Q/C

C1 --> V=(307.8C)/(8F)=38.48V

6.00F --> V=(307.8C)/(6.00F)=51.3V

2.00F--> V=(144/2)=72V

C2 --> V=(144/8)=18V

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects