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184_notes:examples:week8_cap_parallel [2017/10/10 13:44] – created tallpaul | 184_notes:examples:week8_cap_parallel [2017/10/17 00:23] – dmcpadden | ||
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- | =====Example: Application of Ohm's Law===== | + | ===== Connecting Already-Charged Capacitors |
- | Suppose you have a simple circuit that contains only a 9-Volt battery | + | Suppose you have the following setup of already-charged capacitors. The positive plates are all on the top half of the circuit. Capacitors are labeled 1 through 3 for convenience of reference, |
+ | |||
+ | {{ 184_notes: | ||
===Facts=== | ===Facts=== | ||
- | * $\Delta V = 9\text{ | + | * $Q_1 = Q_2 = Q_3 = 1 \text{ |
- | * $R = 120 \Omega$ | + | * $\Delta V_1 = \Delta V_2 = \Delta V_3 = 20 \text{ V}$ |
+ | * Capacitors are aligned as shown in picture. | ||
+ | * Capacitor 2 is flipped in the second case. | ||
===Lacking=== | ===Lacking=== | ||
- | * Current | + | * $C_{\text{equiv}}$ |
+ | * Explanations for what happens after the switches are closed, in both cases. | ||
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The wire has very very small resistance when compared to the 120 $\Omega$ resistor. | + | * The potential differences across the segments of wire are very very small in comparison to the potential differences across |
- | * The circuit is in a steady state. | + | |
- | * Approximating | + | |
- | * There is no outside influence on the circuit. | + | |
===Representations=== | ===Representations=== | ||
- | * We represent [[184_notes: | + | * We represent |
- | * We represent the situation with following circuit diagram. | + | * We represent the [[184_notes: |
- | + | $$\Delta | |
- | {{ 184_notes: | + | * We represent the equivalent capacitance of multiple capacitors arranged in parallel as |
+ | $$C_{\text{equiv}} = C_1 + C_2 + C_3 + \ldots$$ | ||
+ | * We represent the situation with diagram given above. | ||
====Solution==== | ====Solution==== | ||
- | We have assumed that the battery | + | === Part 1 === |
+ | All the charges | ||
+ | $$C_1=C_2=C_3= \frac{Q}{\Delta V} = 50 \mu\text{F}$$ | ||
+ | Now, we can find the equivalent capacitance from Node A to Node B, since the capacitors are arranged in parallel: | ||
+ | $$C_{\text{equiv}} | ||
+ | |||
+ | Okay, so what happens when we closed all the switches? Now the capacitors are connected to one another. A good check is to see if charge will move. If there is a potential differences anywhere across the capacitors, this would cause some or all of the charges to move. The notes tell us that the [[184_notes: | ||
+ | |||
+ | ===Part 2=== | ||
+ | {{ 184_notes: | ||
+ | In the case that Capacitor 2 is flipped, we have the setup shown to the right. When we check the potential | ||
+ | |||
+ | If we consider that charge cannot flow across a capacitor, we know that charges can only move through the wires (i.e., top plate to top plate or bottom plate to bottom plate). Initially, the total charge on the top plates is $1 \text{ mC}$ since there is $1 \text{ mC } - 1 \text{ mC }+1 \text{ mC }$ (and the total charge on the bottom plates is then $-1 \text{ |