Capacitors

[mspeedwagon]

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Hi All,
I'm doing some E&M review questions for the MCAT. I'd love if someone can give me a detailed solution for below.
Thanks.

Two capacitors of 6 μF and 8 μF are connected in parallel. The combination is then connected in series with a 12 V battery and 14 μF capacitor. What is the voltage across the 6 μF capacitor?
a) 12 b) 4 c) 6 d) 5.

 

[Jepstein30]

Hi All,
I'm doing some E&M review questions for the MCAT. I'd love if someone can give me a detailed solution for below.
Thanks.

Two capacitors of 6 μF and 8 μF are connected in parallel. The combination is then connected in series with a 12 V battery and 14 μF capacitor. What is the voltage across the 6 μF capacitor?
a) 12 b) 4 c) 6 d) 5.

What's the answer?? If its C, I can explain If not, I'll be checking back in here to learn something!

 

[sciencebooks]

What's the answer?? If its C, I can explain If not, I'll be checking back in here to learn something!

I also got C, but I'm not sure if I didn't it correctly either.

OP, were the answer choices given with units, btw?

 

[mspeedwagon]

C is correct and yes, the choices are all Volts. How did you get C? Help anyone?

I also got C, but I'm not sure if I didn't it correctly either.

OP, were the answer choices given with units, btw?

 

[mspeedwagon]

I keep getting 9 V which isn't an option. I'm sure I'm doing something simple wrong. Would appreciate any input.

 

[indianjatt]

First, you must find the total equivalent capacitance. In this case, it is 6+ 8 = 14 (for the parallel sum). Then we must add the two capacitor equivalence with the other capacitor in series. (1/Ctotal = 1/14 + 1/14 = 2/14 . C total = 7).

Now we have our total equivalence resistance. We can find our total charge. C=Q/V. Q=CV.
Q = 7*12 = 84.

Now, just find the voltage drop across a 14 microfaraday capacitory. C=Q/V. V = Q/C . V = 84/14 = 6V

 

[smiley27]

Hi All,
I'm doing some E&M review questions for the MCAT. I'd love if someone can give me a detailed solution for below.
Thanks.

Two capacitors of 6 μF and 8 μF are connected in parallel. The combination is then connected in series with a 12 V battery and 14 μF capacitor. What is the voltage across the 6 μF capacitor?
a) 12 b) 4 c) 6 d) 5.

The first two parallel capacitors have an equivalent capacitance of 14 uF (6+8, parallel). This equivalent 14 uF capacitor is now in series with the third 14 uF capacitor. The total equivalent capacitance for all three capacitors is 7 uF.

Q=CeqV=7uF x 12V = 84 micro coulombs. 84 microcoulombs which is the charge on all of the capacitors is used to find the voltage drop across the first two capacitors using the equivalent capacitance of the two parallel capacitors.
Q=CeqV:
84 microcoulombs = 14 uF x V drop across just the two parallel capacitors. Solving for V you get a drop of 6V across the parallel combo but they're in parallel so both of the capacitors receive the same V drop. 6V

 

[mspeedwagon]

Thanks... was close... go to Q = 84, but got stuck.