Circuit A: $$R_{eq}=R_1+R_2+R_3= 9 Ω$$ $$V_3>V_2>V_1$$ $$V=IR$$ $$I_1=I_2=I_3$$

Circuit B: $$C_{eq}=(\frac{1}{C_1}+ \frac{1}{C_2}+ \frac{1}{C_3})^{-1}= 2.3 mF$$ $$Q_{1}=Q_{2}=Q_{3}$$ $$Q=CV$$ $$V_1>V_2>V_3$$

Circuit C: $$C_{eq}=C_1+C_2+C_3 = 21 mF$$ $$V_{1}=V_{2}=V_{3}$$ $$Q=CV$$ $$Q_3>Q_2>Q_1$$

Circuit D: $$R_{eq}=(\frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3})^{-1}= 0.92 Ω$$ $$V_{1}=V_{2}=V_{3}$$ $$I_1>I_2>I_3$$

Circuit A: $$R_{2}||R_{4}||R_{5}$$ $$R_{245}=(\frac{1}{R_2}+ \frac{1}{R_4}+ \frac{1}{R_5})^{-1}$$ $$R_{1}→R_{245}→R_{3}$$ $$R_{eq}=R_1+R_{245}+R_3$$ $$R_{eq}= 350 Ω$$

Circuit B: $$R_{2}→R_{3}$$ $$R_{23}=R_2+R_3$$ $$R_{23}||R_5$$ $$R_{235}=(\frac{1}{R_{23}}+ \frac{1}{R_5})^{-1}$$ $$R_{1}→R_{235}→R_{4}$$ $$R_{eq}=R_1+R_{235}+R_4$$ $$R_{eq}= 400 Ω$$

Circuit C: $$C_{1}||C_{2}, C_{3}||C_{4}$$ $$C_{12}=C_1+C_2, C_{34}=C_3+C_4$$ $$C_{12}→C_{5}→C_{34}$$ $$C_{eq}=(\frac{1}{C_{12}}+ \frac{1}{C_5}+\frac{1}{C_{34}})^{-1}$$ $$C_{eq}=235 μF$$

Circuit D: $$C_{4}→C_{5}$$ $$C_{45}=(\frac{1}{C_{2}}+ \frac{1}{C_{45}})^{-1}$$ $$C_{2}||C_{45}$$ $$C_{245}=C_2+C_{45}$$ $$C_{1}→C_{245}→C_{3}$$ $$C_{eq}=(\frac{1}{C_1}+ \frac{1}{C_{245}}+\frac{1}{C_3})^{-1}$$ $$C_{eq}=176.25 μF$$

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:

1. $V_1=I_1R_1$
2. $I_1=I_2$ b/c in series
3. $V_{bat}=V_1+V_2+V_5$ loop rule, find $V_2$
4. $V_3+V_4=V_5$ loop rule, find $V_4$
5. $I_4=I_3$ b/c in series
6. $I_1=I_3+I_5$ b/c junction, find $I_5$
7. Solve for the rest with $V=IR$ and $P=IV$

Results:

Power Ranking: $P_1=P_2>P_1=P_3=P_4$

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:

1. $I_4=I_2$ b/c in series
2. $I_3=I_5$ b/c in series
3. $V_{bat}=V_1$ b/c parallel
4. $V_{bat}=V_2+V_4$ b/c loop rule
5. $V_{bat}=V_3+V_5$ b/c loop rule
6. $P_1=I_1V_1$ find $I_1$
7. Solve for the rest with $V=IR$ and $P=IV$
8. $I_{bat}=I_1+I_2+I_3$ b/c junction

Results:

Power Ranking: $P_1>P_5>P_2>P_3>P_4$

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:

1. $V_1=V_2$ b/c in parallel
2. $P_1=I_1V_1$ find $I_1$
3. $I_{tot}=I_3=I_1+I_2$ b/c junction
4. $P_5=I_5V_5$ find $V_5$
5. $V_5=V_4$ b/c in parallel
6. $I_4=I_{tot}-I_5$ b/c junction
7. $V_{bat}=V_1+V_3+V_4$ b/c loop rule
8. Solve for the rest with $V=IR$ and $P=IV$

Results:

Power Ranking: $P_5>P_3>P_1=P_2>P_4$

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:

1. $I_{bat}=I_5=I_4$ b/c in series
2. $P_4=I_4V_4$ find $V_4$
3. $P_5=I_5V_5$ find $V_5$
4. $V_{bat}=V_1+V_4+V_5$ b/c loop rule
5. $V_{bat}=V_2+V_3+V_4+V_5$ b/c loop rule
6. $P_1=I_1V_1$ find $I_1$
7. $I_{bat}=I_1+I_2$ b/c junction rule
8. $I_2=I_3$ b/c series
9. Solve for the rest with $V=IR$ and $P=IV$

Results:

Power Ranking: $P_1>P_4>P_5>P_3>P_2$

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:

1. $V_1=V_{bat}-V_2$
2. $V_5=V_{bat}-V_1-V_4$
3. $V_3=V_2$
4. $U_2=\frac{1}{2}Q_2V_2$, find $Q_2$
5. $Q_1=Q_2+Q_3+Q_4$
6. $Q_4=Q_5$
7. Solve for the rest with $U=\frac{1}{2}QV$ and $Q=CV$

Results:

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:

1. Solve for $V_1$, $U_1=\frac{1}{2}Q_1V_1$
2. $V_3= V_1-V_2$
3. $Q_5=Q_4$
4. Solve for $V_4$, $U_4=\frac{1}{2}Q_4V_4$
5. $V_3=V_4+V_5$
6. $Q_2=Q_3=Q_5$
7. Solve for the rest with $U=\frac{1}{2}QV$ and $Q=CV$

Results:

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:

1. V_{bat}=V_1+V_2+V_3
2. $V_4$= V_5$ - $Solve for $Q_4$, $U_4=\frac{1}{2}Q_4V_4$
3. $V_2=V_3+V_5$
4. Solve for $Q_2$, $U_2=\frac{1}{2}Q_2V_2$
5. $Q_5=Q_4+Q_2$
6. $Q_3=Q_4+Q_5$
7. Solve for the rest with $U=\frac{1}{2}QV$ and $Q=CV$

Results:

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:

1. $V_4=V_5$
2. Solve for $Q_5$, $U_5=\frac{1}{2}Q_5V_5$
3. Solve for $V_1$, $U_1=\frac{1}{2}Q_1V_1$
4. $V_1+V_2=V_3+V_4$
5. Solve for $Q_3$, $U_{34}=\frac{1}{2}Q_3V_3$
6. $Q_3=Q_4+Q_5$
7. $Q_1=Q_2$
8. Solve for the rest with $U=\frac{1}{2}QV$ and $Q=CV$

Results:

Given:

$V_1= 9V$, $V_2= 6V$, $R=100 Ω$

Simplify Circuit:

• $R_1$ and $R_2$ in series, $R_1+R_2=200 Ω$

Node Rule:

• $I_1+I_2=I_3$

Loop Rule:

• Loop A: $V_1-I_1R_{12}-I_3R_3=0$
• Loop B: $I_3R_3-I_2R_4-V_2=0$
• Loop C: $V_1-I_1R_{12}-I_2R_4-V_2=0$

Solution:

$I_1=0.024A$, $I_2=0.018A$, $I_3=0.042A$

*note can use wolfram/online/calc to evaluate I from loop AND node rule equations

Given:

$V_1= 9V$, $V_2= 6V$, $R=100 Ω$

Simplify Circuit:

• $R_1$ || $R_2$, $R_{12}=50 Ω$

Node Rule:

• $I_1=I_2+I_3$

Loop Rule:

• Loop A: $V_1-I_1R_{12}-I_3R_3=0$
• Loop B: $ V_2 -I_2R_4 +I_3R_3=0$
• Loop C: $V_1-I_1R_{12-V_2}-I_2R_4=0$

Solution:

$I_1=0.12 A$, $I_2=0.09A$, $I_3=0.03A$

Given:

$V_1= 9V, V_2= 6V, R=100 Ω$

Simplify Circuit:

• $R_2$ and $R_3$ in series, $R_2+R_3=200 Ω$
• $R_4$ || $R_5$, $R_{12}=50 Ω$

Node Rule:

• $I_1=I_2+I_3$

Loop Rule:

• Loop A: $V_1-I_1R_1+V_2-I_1R_{45}=0$
• Loop B: $ -V_2-I_3R_{23}=0$
• Loop C: $V_1-I_1R_1-I_3R_{23}-I_1R_{45}=0$

Solution:

$I_1=0.06A$, $I_2=0.09A$, $I_3=-0.03A$

Given:

$V_1= 9V, V_2= 6V, R=100 Ω$

Simplify Circuit:

• $R_2$ and $R_3$ in series, $R_2+R_3=200 Ω$
• $R_1$||$R_{23}$, $R_{12}=66.667 Ω$

Node Rule:

• $I_1=I_2+I_3$

Loop Rule:

• Loop A: $V_1-I_1R_{123}- 0$
• Loop B: $ I_2R_{123}-V_2-I_3R_4=0$
• Loop C: $V_1-V_2-I_3R_4=0$

Solution:

$I_1=0.165A$, $I_2=0.135A$, $I_3=0.03A$

1. Redraw the circuit
2. Look for easy combinations of circuit elements
3. Pick the direction of current in each branch of the circuit (LABEL CLEARLY AND DO NOT CHANGE!!!!!!!!!)
4. Identify the Nodes and write out the Node Rule equations
5. Identify the Loops and write out the Loop Rule equations
6. 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).
7. Solve the system of equations (you can use Wolfram Alpha or other resources online to solve the system of equations)