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184_notes:examples:week8_cap_parallel [2017/10/10 13:44] – created tallpaul | 184_notes:examples:week8_cap_parallel [2017/10/11 18:22] – [Connecting Already-Charged Capacitors] tallpaul | ||
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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$$ | ||
+ | * We represent the situation with diagram given above. The flipped situation is below. | ||
+ | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | We have assumed that the battery | + | All the charges |
+ | $$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 to see if charge will move involves the potential differences across the capacitors. The notes tell us that the [[184_notes: | ||
+ | |||
+ | In the case that Capacitor 2 is flipped, we have the setup shown in the representations list. When we check the potential | ||
+ | |||
+ | If we consider that charge cannot flow across a capacitor, we know that the top half and the bottom half of the setup cannot exchange charge. So the total charge on top is $50 \mu\text{F}$, and the total charge on the bottom is $-50 \mu\text{F}$. In order to reach a steady state, since each capacitor is the same, the charge should spread out evenly among the capacitors. So we end up with each capacitor begin charged to $16.67 \mu\text{F}$. This way, charge is conserved on the top and bottom, and each capacitor is charged to the same amount. The steady state should look the original diagram given, just with less (a third, to be exact) of the charge built up on the plates. |