Chapter+16+Capacitors


 * Chapter 16 Capacitors**-- this page has been edited by avi stien.

THIS WAS MADE BY AVI STEIN

"In a way, a capacitor is a little like a battery. Although they work in completely different ways, capacitors and batteries both **store electrical energy**." - []

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C=Q/∆V where: 1.**C** = constant capacitance (in units Faraday (F)) 2. **Q** = charge (in units coulomb) 3. ∆**V** = potential difference (def:the negative of work done by conservative force)

voltage in parallel

voltage in series ∆Vtotal= ∆V1+∆V2+...

this is for capacitors in series



this is for capacitors in parallel



charge in series Qeq=Q1+Q2+...

charge in parallel Qeq=Q1+Q2+...

practice questions! :D

First add (1/6) to (1/3) which would give you (3/6). convert that to (1/2) and then invert it to get 2. the charge in the circuit ends up being 2 faradays.
 * 1. if you have a circuit with two capacitors in series, one with a charge of 6 C and one with a charge of 3 C, what is the capacitance equivalent?**

This is simple. All I do is add them together. 1 + 999 is 1000. this leads me to the very astute conclusion that the charge equivalent is none other than 1000 faradays!
 * 2. There is a circuit with two capacitors in parallel, one with a charge of 999 C and the other with a charge of 1 C, what is the capacitance equivalent?**

well the formula tells me that the voltage in one is equal to the voltage in the other and therefore it would be π v.
 * 3. There is a circuit with a battery** **in parallel with a voltage of π** **and two capacitors. what is the voltage in capacitor A in the circuit?**


 * 4. There is a circuit with two capacitors in series. One has a voltage of 11 and the other has a voltage of 7. What is the total voltage in the battiery to the left of both of the batteries?**

well in order to solve this intriguing quandary it is important to recall that all that is necessary is to simply add the two voltages togeather. 7 +11 is 18 and therefore the total voltage in the battiery is 18v.


 * 5. there is yet another circuit with two capacitors in parallel each with a charge of 444. what is the charge equivalent?**

it is always a pleasure when only simple arithmetic is necessary to solve a dilemma and that is why this is such a delight. we simply add 444 to 444 to garner a grand total of 888.


 * 6. on the last and final circuit there is a charge of 1029 q received at the convergence of two capacitors in series. the observer, being Curious George in the flesh does the very unusual, and wants to know something intelligent. In fact he asks me what the charge is in each individual capacitor. how should I answer him?**

well since i know that each of the capacitors has the same charge as the charge equivalent and I know that since the charge equivalent is 1029, each capacitor must have a charge of 1029 also. brilliant!

7. if i have a measly old charge of 16q and a very nice potential difference of 4 what will my constant capacitance be in faradays? well its simple. all i do is devide 16 by 4. well thats a no-brainer, it's obvouly 4. the potential difference and capacitance are twinsies.

mutiple choice

1. what will result in a large capacitance? a. large Q and large ∆V b large Q and small ∆v c small q and large ∆V d small q and small ∆v

2. if there are two capacitors each with a capactence of 7 and they are in parallel what is the capacitence equivelent? a. 7 b 14 c 49 d 7/2

3. when there is a capacitance equivalent to 15/8 and one of the capacitors that is in series is 3 what must the other one be? a 5/8 b 45/8 c 45 d 5

4. the charge in both capacitors in a parallel circuit are 4, what is the the equivalent charge? a. 1 b 4 c 8 d 16

5 the total voltage in a curcuit in series is 12. one capacitor is 3x another is 4x and there third is just x. what must x be? a 1.5 b 8 c 7 d 12