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--- 184_notes:examples:week8_cap_parallel [2017/10/11 16:58] – tallpaul
+++ 184_notes:examples:week8_cap_parallel [2018/06/26 14:45] (current) – curdemma
@@ Line 1: / Line 1: @@
+[[184_notes:c_parallel|Return to Capacitors in Parallel notes]]
 ===== 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? Part 2: What if Capacitor 2 were flipped?
-{{ 184_notes:8_cap_parallel.png?300 |Circuit with Capacitors in Parallel}}
+[{{ 184_notes:8_cap_parallel.png?500 |Circuit with Capacitors in Parallel}}]
 ===Facts===
@@ Line 23: / Line 25: @@
   * 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.
+  * We represent the situation with diagram given above.
-{{ 184_notes:8_cap_parallel_flipped.png?300 |Circuit with Capacitors in Parallel}}
 ====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 (this is not always the case for capacitors in parallel, but just happens to be for this example):
  $C_1=C_2=C_3= \frac{Q}{\Delta V} = 50 \mu\text{F}$
 Now, we can find the equivalent capacitance from Node A to Node B, since the capacitors are arranged in parallel:
  $C_{\text{equiv}} = C_1 + C_2 + C_3 = 150 \mu\text{F}$
-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:c_parallel#Loop_Rule_and_Voltage_in_Parallel|potential differences across parallel capacitors]] will be the same in a steady state. If they are not the same, we can use the Loop Rule to show there is a voltage difference along the wire, which will cause charge to flow between capacitors. We are happy to see that the potential differences across our capacitors are the same, so the setup is already in a steady state! Nothing changes when the switches are closed.
+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:c_parallel#Loop_Rule_and_Voltage_in_Parallel|potential differences across parallel capacitors]] will be the same in a steady state. If they are not the same, we can use the Loop Rule to show there is a voltage difference along the wire, which will cause charge to flow between capacitors. We are happy to see that the potential differences across our capacitors are the same, so the setup is already in a steady state! **Nothing changes when the switches are closed.**
-In the case that Capacitor 2 is flipped, we have the setup shown in the representations list. 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 2 and go through  $\Delta V_2=-12 \text{ V}$ , or through one of the other capacitors and go through  $\Delta V_{1,3}=+12 \text{ V}$ . These are different, but that can't be! There must be some potential difference in the wire that we are not accounting for. This potential difference will cause charge to flow. The only question is how.
-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  $-50 \mu\text{F}$ .
-We use  $I_1$  since this is the current in the wire connected directly to the battery. We can break this down further to find the equivalent resistance in the chunk of the circuit containing Resistors 2, 3, and 4. This chunk and Resistor 1 are connected in series to form the resistance of the whole circuit, so we can use equation (2) to write:
- $R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, chunk with 2,3,4}}$
-This yields  $R_{\text{equiv, chunk with 2,3,4}}=160 \Omega$ .
-Notice that this "chunk" is actually two parallel pieces of the circuit, starting at Node A and ending at Node B. The two parallel parts are Resistor 2, and then Resistors 3 and 4 together. We can use equation (3) to break down the chunk into these two pieces (we combine Resistors 3 and 4 below using equation (2)):
- $\frac{1}{R_{\text{equiv, chunk with 2,3,4}}}= \frac{1}{R_2}+\frac{1}{R_3+R_4}$
-We can plug in what we know and solve for the resistance of Resistor 2:
- $R_2=200\Omega$
-Okay, now for the potential differences. It will be useful in the approach we choose to know the current through Resistor 4, which is found from Ohm's Law:
- $I_4=\frac{\Delta V_4}{R_4}=10 \text{ mA}$
-A simple application of Node Rule -- equation (5) -- at Node C should tell us that  $I_3 = I_4$ . Now, we can reapply Ohm's Law to find the potential difference across Resistor 3:
- $\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$
-{{ 184_notes:8_res_parallel_loop_1.png?300 |Circuit with Resistors in Series and Parallel}}
+===Part 2===
-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), voltage decreases, so  $\Delta V_1$ ,  $\Delta V_3$ ,  $\Delta V_4<0$ , whereas we have $\Delta V_{\text{bat}}>0$. These four potential differences form a loop, so they should add to 0:
+[{{  184_notes:8_cap_parallel_flipped.png?500|Circuit with Capacitors in Parallel}}]
-$$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$
+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 2 and go through $\Delta V_2=-12 \text{ V}$, or through one of the other capacitors and go through $\Delta V_{1,3}=+12 \text{ V}$. These are different, but that can't be! There must be some potential difference in the wire that we are not accounting for. This potential difference will cause charge to flow. The only question is how.
-We know enough potential differences to find the voltage across Resistor 1:
-$$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ V}$$
-{{ 184_notes:8_res_parallel_loop_2.png?300 |Circuit with Resistors in Series and Parallel}}
+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 $-1 \text{ mC}$). In order to reach a steady state, since each capacitor is the same, the charge should spread out evenly among the capacitors because they are connected by conductors. So each capacitor would have a charge of  $0.33 \text{ mC}$  after returning to equilibrium. This way, charge is conserved on the top and bottom, and each capacitor is charged to the same amount (though this happens because the capacitors all have the same capacitance, not because they are in parallel). 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.
-Now, consider the (different!) loop highlighted in the circuit above. The Loop Rule tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. We see that current in Resistor 2 runs opposite to the current in the other resistors if we follow the loop in one direction. We have to choose a direction for the application of the Loop Rule. If we go clockwise, voltage increases across Resistor 2, but drops across Resistors 3 and 4. So we write:
-$$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$
-We know enough potential differences to find the voltage across Resistor 2:
-$$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ V}$$
-That's all! Note that there are a lot of ways to do this problem, but we chose an approach that showcases the power of knowing equivalent resistance for resistors in parallel, and the power of the Loop Rule. See if you can create a different method for finding the unknowns.