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184_notes:examples:week8_cap_parallel [2017/10/11 15:53] – tallpaul | 184_notes:examples:week8_cap_parallel [2017/10/17 00:23] – dmcpadden | ||
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- | ===== Example: Resistors in Series and in Parallel | + | ===== Connecting Already-Charged Capacitors |
- | Suppose you have the following circuit. | + | Suppose you have the following |
- | {{ 184_notes:8_res_parallel.png?300 |Circuit with Resistors | + | {{ 184_notes:8_cap_parallel.png?500 |Circuit with Capacitors |
===Facts=== | ===Facts=== | ||
- | * $\Delta V_{\text{bat}} | + | * $Q_1 = Q_2 = Q_3 = 1 \text{ |
- | * $I_1 = 50 \text{ | + | * $\Delta V_1 = \Delta V_2 = \Delta |
- | * $R_1=80 | + | * Capacitors are aligned as shown in picture. |
- | * $R_3=300 \Omega$ | + | * Capacitor 2 is flipped in the second case. |
- | * $R_4=500 \Omega$ | + | |
- | * $\Delta | + | |
===Lacking=== | ===Lacking=== | ||
- | * $\Delta V_1$, $\Delta V_2$, $\Delta V_3$, $R_2$. | + | * $C_{\text{equiv}}$ |
+ | * 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 | + | * The potential differences across the segments of wire are very very small in comparison |
- | * The circuit is in a steady state. | + | |
- | * Approximating the battery as a mechanical battery. | + | |
- | * The resistors in the circuit are made of Ohmic materials. | + | |
===Representations=== | ===Representations=== | ||
- | * We represent | + | * We represent |
- | \begin{align*} | + | |
- | \Delta V = IR &&&&&& | + | |
- | \end{align*} | + | |
- | * We represent the equivalent resistance of multiple resistors arranged in series as | + | |
- | \begin{align*} | + | |
- | R_{\text{equiv, | + | |
- | \end{align*} | + | |
- | * We represent the equivalent resistance of multiple resistors arranged in parallel as | + | |
- | \begin{align*} | + | |
- | \frac{1}{R_{\text{equiv, parallel}}}= \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots &&&&&& | + | |
- | \end{align*} | + | |
* We represent the [[184_notes: | * We represent the [[184_notes: | ||
- | \begin{align*} | + | $$\Delta V_1+\Delta V_2+\Delta V_3+\ldots = 0$$ |
- | \Delta V_1+\Delta V_2+\Delta V_3+\ldots = 0 &&&&&& | + | * We represent the equivalent capacitance of multiple capacitors arranged |
- | \end{align*} | + | $$C_{\text{equiv}} = C_1 + C_2 + C_3 + \ldots$$ |
- | * We represent the [[184_notes: | + | |
- | \begin{align*} | + | |
- | I_{\text{in}} = I_{\text{out}} &&&&&& | + | |
- | \end{align*} | + | |
* We represent the situation with diagram given above. | * We represent the situation with diagram given above. | ||
====Solution==== | ====Solution==== | ||
- | Let's start with resistance. The equivalent resistance for the entire circuit can be found with Ohm's Law -- equation | + | === Part 1 === |
- | $$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$ | + | All the charges and potential differences across the capacitors are the same, so they should have the same capacitance |
+ | $$C_1=C_2=C_3= | ||
+ | Now, we can find the equivalent capacitance from Node A to Node B, since the capacitors are arranged in parallel: | ||
+ | $$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 | + |