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--- 184_notes:level_up_sol [2020/10/26 21:14] – dmcpadden
+++ 184_notes:level_up_sol [2020/10/26 21:41] (current) – dmcpadden
@@ Line 26: / Line 26: @@
 ====Level 1====
-Circuit A:
+Circuit A:  $R_{eq}= 350 Ω$
- $R_{2}||R_{4}||R_{5}$
- $R_{245}=(\frac{1}{R_2}+ \frac{1}{R_4}+ \frac{1}{R_5})^{-1}$
- $R_{1}→R_{245}→R_{3}$
- $R_{eq}=R_1+R_{245}+R_3$
-$$R_{eq}= 350 Ω$$
-Circuit B:
+Circuit B:  $R_{eq}= 400 Ω$
- $R_{2}→R_{3}$
- $R_{23}=R_2+R_3$
- $R_{23}||R_5$
- $R_{235}=(\frac{1}{R_{23}}+ \frac{1}{R_5})^{-1}$
- $R_{1}→R_{235}→R_{4}$
- $R_{eq}=R_1+R_{235}+R_4$
-$$R_{eq}= 400 Ω$$
+Circuit C:  $C_{eq}=235 μF$
+Circuit D:  $C_{eq}=176.25 μF$
-Circuit C:
- $C_{1}||C_{2}, C_{3}||C_{4}$
- $C_{12}=C_1+C_2, C_{34}=C_3+C_4$
- $C_{12}→C_{5}→C_{34}$
- $C_{eq}=(\frac{1}{C_{12}}+ \frac{1}{C_5}+\frac{1}{C_{34}})^{-1}$
- $C_{eq}=235 μF$
-Circuit D:
- $C_{4}→C_{5}$
- $C_{45}=(\frac{1}{C_{2}}+ \frac{1}{C_{45}})^{-1}$
- $C_{2}||C_{45}$
- $C_{245}=C_2+C_{45}$
- $C_{1}→C_{245}→C_{3}$
- $C_{eq}=(\frac{1}{C_1}+ \frac{1}{C_{245}}+\frac{1}{C_3})^{-1}$
-$$C_{eq}=176.25 μF$$
 ====Level 2====
 ===Circuit A:===
-Simplify Circuit:
-  *  $R_3$  and  $R_4$  in series
-  *  $R_{34}||R_5$
-  *  $R_1$ ,  $R_2$ , and  $R_{345}$  in series
-  *  $R_{eq}=\frac{V_{bat}}{I_{bat}}=1.167 Ω$
-Steps:
-  -  $V_1=I_1R_1$
-  -  $I_1=I_2$  b/c in series
-  -  $V_{bat}=V_1+V_2+V_5$   loop rule, find  $V_2$
-  -  $V_3+V_4=V_5$   loop rule, find  $V_4$
-  -  $I_4=I_3$  b/c in series
-  -  $I_1=I_3+I_5$  b/c junction, find  $I_5$
-  - Solve for the rest with  $V=IR$  and  $P=IV$
 Results:
 {{:course_planning_studio_em:week8:2a.png?400|}}
 Power Ranking:
@@ Line 91: / Line 45: @@
 ===Circuit B:===
-Simplify Circuit:
-  *  $R_2$  and  $R_4$  in series
-  *  $R_3$  and  $R_5$  in series
-  *  $R_1$ || $R_{24}$ || $R_{35}$
-  *  $R_{eq}=\frac{V_{bat}}{I_{bat}}=2 Ω$
-Steps:
-  -  $I_4=I_2$  b/c in series
-  -  $I_3=I_5$  b/c in series
-  -  $V_{bat}=V_1$  b/c parallel
-  -  $V_{bat}=V_2+V_4$  b/c loop rule
-  -  $V_{bat}=V_3+V_5$  b/c loop rule
-  -  $P_1=I_1V_1$  find  $I_1$
-  - Solve for the rest with  $V=IR$  and  $P=IV$
-  -  $I_{bat}=I_1+I_2+I_3$  b/c junction
 Results:
@@ Line 116: / Line 53: @@
 ===Circuit C===
-Simplify Circuit:
-  *  $R_1$ || $R_2$
-  *  $R_5$ || $R_4$
-  *  $R_{12}$ ,  $R_3$ , and  $R_{45}$  in series
-  *  $R_{eq}=\frac{V_{bat}}{I_{bat}}=1.42 Ω$
-Steps:
-  -  $V_1=V_2$  b/c in parallel
-  -  $P_1=I_1V_1$  find  $I_1$
-  -  $I_{tot}=I_3=I_1+I_2$  b/c junction
-  -  $P_5=I_5V_5$  find  $V_5$
-  -  $V_5=V_4$  b/c in parallel
-  -  $I_4=I_{tot}-I_5$  b/c junction
-  -  $V_{bat}=V_1+V_3+V_4$  b/c loop rule
-  - Solve for the rest with  $V=IR$  and  $P=IV$
 Results:
@@ Line 142: / Line 62: @@
 ===Circuit D===
-Simplify Circuit:
-  *  $R_2$  and  $R_3$  in series
-  *  $R_2$ || $R_{23}$
-  *  $R_{123}$ ,  $R_4$ , and  $R_5$  in series
-  *  $R_{eq}=\frac{V_{bat}}{I_{bat}}=4.75Ω$
-Steps:
-  -  $I_{bat}=I_5=I_4$  b/c in series
-  -  $P_4=I_4V_4$  find  $V_4$
-  -  $P_5=I_5V_5$  find  $V_5$
-  -  $V_{bat}=V_1+V_4+V_5$  b/c loop rule
-  -  $V_{bat}=V_2+V_3+V_4+V_5$  b/c loop rule
-  -  $P_1=I_1V_1$  find  $I_1$
-  -  $I_{bat}=I_1+I_2$  b/c junction rule
-  -  $I_2=I_3$  b/c series
-  - Solve for the rest with  $V=IR$  and  $P=IV$
 Results:
@@ Line 169: / Line 71: @@
 ====Level 3====
 ===Circuit A===
-Simplify Circuit:
-  *  $C_4$  and  $C_5$  in series
-      *  $C_{45}=(\frac{1}{C_4}+ \frac{1}{C_5})^{-1}=75uF$
-  *  $C_{45}$ || $C_3$ || $C_2$
-      *  $C_{2345}=C_{45} + C_3+C_2=645uF$
-  *  $C_{2345}$  and  $C_1$  in series
-      *  $C_{tot}=(\frac{1}{C_1}+ \frac{1}{C_{2345}})^{-1}=204.76uF$
-  *  $Q_{tot}=C_{tot}V_{tot}$
-      *  $Q_{tot}=(204.76*10^{-6})(9)=1.84mC$
-      *  $Q_{tot}=Q_1=Q_{2345}$
-  *  $Q_{2345}= C_{2345} V_{2345}$
-      *  $V_{2345}=V_2=V_3=V_{45}=2.867V$
-  *  $V_{45}C_{45}=Q_{45}$
-      *  $Q_{45}=Q_4=Q_5=0.215mC$
-  *  $U=\frac{1}{2}QV$  for all
-Steps:
-  -  $V_1=V_{bat}-V_2$
-  -  $V_5=V_{bat}-V_1-V_4$
-  -  $V_3=V_2$
-  -  $U_2=\frac{1}{2}Q_2V_2$ , find  $Q_2$
-  -  $Q_1=Q_2+Q_3+Q_4$
-  -  $Q_4=Q_5$
-  - Solve for the rest with  $U=\frac{1}{2}QV$  and  $Q=CV$
 Results:
 {{:course_planning_studio_em:week8:3a.png?400|}}
 ===Circuit B===
-Simplify Circuit:
-  *  $C_4$  and  $C_5$  in series
-      *  $C_{45}=(\frac{1}{C_4}+ \frac{1}{C_5})^{-1}= 6mF$
-  *  $C_{45}$ || $C_3$
-    *   $C_{345}=C_{45} + C_3= 26mF$
-  *  $C_{345}$  and  $C_2$  in series
-      *  $C_{2345}=(\frac{1}{C_2}+ \frac{1}{C_{345}})^{-1}= 7.22mF$
-  *  $C_{2345}$ || $C_1$
-      *  $C_{tot}= C_{2345}+ C_1= 8.22mF$
-  *  $V_{bat}=V_1=V_{2345}$
-  *  $Q_{2345}= C_{2345} V_{2345}$
-      *  $Q_{2345}= 16(0.00722)=0.116 C= Q_2=Q_{345}$
-  *  $Q_{345}= C_{345}V_{345}$
-      *  $V_{345}= 4.46 V = V_3=V_{45}$
-  *  $Q_{45}=V_{45}C_{45}$
-      *  $Q_{45}=0.027 C =Q_4=Q_5$
-Steps:
-  - Solve for  $V_1$ ,  $U_1=\frac{1}{2}Q_1V_1$
-  -  $V_3= V_1-V_2$
-  -  $Q_5=Q_4$
-  - Solve for  $V_4$ ,  $U_4=\frac{1}{2}Q_4V_4$
-  -  $V_3=V_4+V_5$
-  -  $Q_2=Q_3=Q_5$
-  - Solve for the rest with  $U=\frac{1}{2}QV$  and  $Q=CV$
 Results:
@@ Line 231: / Line 81: @@
 ===Circuit C===
-Simplify Circuit:
-  *  $C_4$ || $C_5$
-      *  $C_{45}=C_4+ C_5= 57nF$
-  *  $C_{45}$  and  $C_3$  in series
-      *  $C_{345}=(\frac{1}{C_3}+ \frac{1}{C_{45}})^{-1}= 45.3nF$
-  *  $C_{345}$ || $C_2$
-    *   $C_{2345}= C_{345}+ C_2= 145.3nF$
-  *  $C_{2345}$  and  $C_1$  in series
-      *  $C_{tot}=(\frac{1}{C_{2345}}+ \frac{1}{C_1})^{-1}= 59.2nF$
-  *  $Q_{tot}=V_{tot}C{tot}$
-      * $Q_{tot}=1.78*10^{-7} C=Q_1=Q_{2345}
-  *  $Q_{2345}= C_{2345} V_{2345}$
-      *  $V_{2345}=1.22V=V_2=V_{345}$
-  *  $Q_{345}= C_{345}V_{345}$
-      *  $Q_{345}= 5.53*10^{-8} C = Q_3=Q_{45}$
-  *  $Q_{45}=V_{45}C_{45}$
-      *  $V_{45}=0.97 V =V_4=V_5$
-Steps:
-  - V_{bat}=V_1+V_2+V_3
-  -  $V_4$ = V_5$
-  -  $Solve for$ Q_4 $,$ U_4=\frac{1}{2}Q_4V_4$
-  -  $V_2=V_3+V_5$
-  - Solve for  $Q_2$ ,  $U_2=\frac{1}{2}Q_2V_2$
-  -  $Q_5=Q_4+Q_2$
-  -  $Q_3=Q_4+Q_5$
-  - Solve for the rest with  $U=\frac{1}{2}QV$  and  $Q=CV$
 Results:
@@ Line 264: / Line 86: @@
 ===Circuit D===
-Simplify Circuit:
-  *  $C_4$ || $C_5$
-      *  $C_{45}=C_4+ C_5= 32mF$
-  *  $C_{45}$  and  $C_3$  in series
-      *  $C_{345}=(\frac{1}{C_3}+ \frac{1}{C_{45}})^{-1}= 19mF$
-  *  $C_1$  and  $C_2$  in series
-      *  $C_{12}=(\frac{1}{C_1}+ \frac{1}{C_2})^{-1}= 6.88mF$
-  *  $C_{12}$ || $C_{345}$
-      *  $C_{tot}= C_{12}+ C_{345}= 25.88mF$
-  *  $C_{345}V_{345}=Q_{345}$
-      *  $Q_{345}= 0.095 C = Q_3=Q_{45}$
-  *  $Q_{45}=C_{45}V_{45}$
-      *  $V_{45}=2.97V =V_4=V_5$
-  *  $Q_{tot}=V_{tot}C{tot}$
-      *  $Q_{tot}= 0.13C$
-      *  $V_{tot}=V_{12}=V_{345}$
-  *  $C_{12}V_{12}=Q_{12}$
-    *   $Q_{12}=0.034 C=Q_1=Q_2$
-Steps:
-  -  $V_4=V_5$
-  - Solve for  $Q_5$ ,  $U_5=\frac{1}{2}Q_5V_5$
-  - Solve for  $V_1$ ,  $U_1=\frac{1}{2}Q_1V_1$
-  -  $V_1+V_2=V_3+V_4$
-  - Solve for  $Q_3$ ,  $U_{34}=\frac{1}{2}Q_3V_3$
-  -  $Q_3=Q_4+Q_5$
-  -  $Q_1=Q_2$
-  - Solve for the rest with  $U=\frac{1}{2}QV$  and  $Q=CV$
 Results:
@@ Line 396: / Line 188: @@
  $I_1=0.165A$ ,  $I_2=0.135A$ ,  $I_3=0.03A$
+====Level Bonus====
+a) Initially there is current in all branches of the circuit (uncharged capacitors act like wires - current can pass through).
-====Board Meeting====
+b) $I_i = 0.00436 A$
-===General Steps When Analyzing Circuits===
-  - Redraw the circuit
-  - Look for easy combinations of circuit elements
-  - Pick the direction of current in each branch of the circuit (LABEL CLEARLY AND DO NOT CHANGE!!!!!!!!!)
-  - Identify the Nodes and write out the Node Rule equations
-  - Identify the Loops and write out the Loop Rule equations
-  - Pick the same number of equations as unknowns in your circuit (Note: at least one equation must be a node equation and at least one equation must be a loop equation. If you pick only nodes or only loops, you will always end up with a 0 = 0 situation at the end, which is technically true but not useful).
-  - Solve the system of equations (you can use  Wolfram  Alpha  or  other resources online to solve the system of equations)
-<WRAP tip>
-=== Discussion Prompts ===
-  * **Question:** How did you set up your loops? How do you know when $\Delta V$ is (+) or (-)?
-  * **Answer:** Have them walk you through one of their loops. If moving from (-) to (+) on a battery  $\Delta V$ -(+) if moving from (+) to (-) then  $\Delta V$ -(-). If moving with the current across a resistor,  $\Delta V_R$  = (-); if moving against the current with a resistor,  $\Delta V_R$  = (+)
-  * **Question:** What process did you take to solve this problem?
-  * **Answer:** Have students go over the steps they took in problem solving here (see general steps above)
-=== Evaluation Questions ===
+c) {{ 184_notes:charginggraphs.png?400 }}
-  * **Question:** How did you simplify your answer? What did these assumptions let you do?
-  * **Answer:**
-    * Steady state current
-    * Perfect batteries
-    * No voltage drop across the wire
-    * These all basically just made the math a lot easier. We didn't have to worry about things changing with time, the battery gradually dying and leaving us with less remaining voltage, or voltage drops across the wires, which would entail us having to know a lot more about the circuit than we do here.
-  * **Question:** If the resistors were replaced with lightbulbs, why would brightness be related to power?
-  * **Answer:** The more energy transferred at a particular time the brighter the light will become. For the filament, high resistance and current is wanted to most effectively dissipate energy (In the form of photons: our warning light). The filament will get hot as energy is dissipated over it, that heat will eventually cause the metal to glow. After a long enough time (which is very short in human time scales), the metal will light up due to the heat. Energy is transferred from collisions between the electrons in the metal into heat.
-  * **Question:** What conservation law does the junction rule relate to?
-  * **Answer:** Conservation of charge
-  * **Question:** What conservation law does the loop rule relate to?
+d) Current goes through all branches without a capacitor (charge capacitors act like a break in the circuit - no current)
-  * **Answer:** Conservation of energy
-  * **Question:** How did you determine what resistors were in series and which were in parallel?
+{{  184_notes:chargedcurrent.png?350  }}
-  * **Answer:** Students should walk through similar logic as above. Saying that two resistors "look" like they're in series or in parallel is not a valid reasoning
-=== Extension Questions ===
+e) $I_f = 0.0036 A$
-  * **Question:**
-  * **Answer:**
-</WRAP>
+f) If the switch is opened, the capacitors would discharge through the resistors below.
+{{  184_notes:dischargecurrentpath.png?350  }}