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184_notes:examples:week8_cap_parallel [2017/10/11 16:24] – [Solution] tallpaul | 184_notes:examples:week8_cap_parallel [2017/10/17 00:23] – dmcpadden | ||
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===== Connecting Already-Charged Capacitors ===== | ===== Connecting Already-Charged Capacitors ===== | ||
- | 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, and the sign of the charge on the plates is indicated. You know that $Q_1 = Q_2 = Q_3 = 1 \text{ mC}$, and $\Delta V_1 = \Delta V_2 = \Delta V_3 = 20 \text{ V}$. What is the equivalent capacitance (if the switches are closed) from Node A to Node B? What happens after the switches are closed? What if Capacitor 2 were flipped? | + | 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, and the sign of the charge on the plates is indicated. You know that $Q_1 = Q_2 = Q_3 = 1 \text{ mC}$, and $\Delta V_1 = \Delta V_2 = \Delta V_3 = 20 \text{ V}$. Part 1: What is the equivalent capacitance (if the switches are closed) from Node A to Node B? What happens after the switches are closed? |
- | {{ 184_notes: | + | {{ 184_notes: |
===Facts=== | ===Facts=== | ||
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* We represent the equivalent capacitance of multiple capacitors arranged in parallel as | * We represent the equivalent capacitance of multiple capacitors arranged in parallel as | ||
$$C_{\text{equiv}} = C_1 + C_2 + C_3 + \ldots$$ | $$C_{\text{equiv}} = C_1 + C_2 + C_3 + \ldots$$ | ||
- | * We represent the situation with diagram given above. The flipped situation is below. | + | * We represent the situation with diagram given above. |
- | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | All the charges and potential differences across the capacitors are the same, so they should have the same capacitance: | + | === Part 1 === |
+ | All the charges and potential differences across the capacitors are the same, so they should have the same capacitance | ||
$$C_1=C_2=C_3= \frac{Q}{\Delta V} = 50 \mu\text{F}$$ | $$C_1=C_2=C_3= \frac{Q}{\Delta V} = 50 \mu\text{F}$$ | ||
Now, we can find the equivalent capacitance from Node A to Node B, since the capacitors are arranged in parallel: | Now, we can find the equivalent capacitance from Node A to Node B, since the capacitors are arranged in parallel: | ||
+ | $$C_{\text{equiv}} = C_1 + C_2 + C_3 = 150 \mu\text{F}$$ | ||
- | We use $I_1$ since this is the current in the wire connected | + | 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 |
- | $$R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, chunk with 2,3,4}}$$ | + | |
- | This yields $R_{\text{equiv, | + | |
- | Notice | + | ===Part 2=== |
- | $$\frac{1}{R_{\text{equiv, chunk with 2,3, | + | {{ 184_notes: |
- | We can plug in what we know and solve for the resistance of Resistor 2: | + | In the case that Capacitor 2 is flipped, we have the setup shown to the right. When we check the potential differences in the different capacitors, we notice that the voltage across Capacitor 2 is the opposite as the voltage across the other capacitors. If we travel from Node B to Node A, we could travel through Capacitor |
- | $$R_2=200\Omega$$ | + | |
- | Okay, now for the potential differences. It will be useful in the approach | + | If we consider that charge cannot flow across a capacitor, we know that charges can only move through |
- | $$I_4=\frac{\Delta V_4}{R_4}=10 \text{ mA}$$ | + | |
- | A simple application of Node Rule -- equation | + | |
- | $$\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$$ | + | |
- | + | ||
- | {{ 184_notes: | + | |
- | A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted in the circuit above. The Loop Rule -- equation (4) -- tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. If we travel along the direction of conventional current (clockwise in our representation), | + | |
- | $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ | + | |
- | We know enough potential differences to find the voltage across Resistor | + | |
- | $$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ | + | |
- | + | ||
- | {{ 184_notes: | + | |
- | Now, consider | + | |
- | $$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$ | + | |
- | We know enough potential differences to find the voltage across Resistor 2: | + | |
- | $$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ | + | |
- | That's all! Note that there are a lot of ways to do this problem, but we chose an approach that showcases | + |