Complex Circuit Problems Worksheet with diagrams and equations for physics students.

To solve the problems in the worksheet, we need to analyze each circuit step by step using fundamental principles of circuit analysis, such as Ohm's Law, Kirchhoff's Laws, and series/parallel resistor combinations. Let's go through each circuit one by one.

---

Circuit 1:

#### Given:
- Resistors: \( R_1 = 8 \Omega \), \( R_2 = 12 \Omega \), \( R_3 = 6 \Omega \)
- Current: \( I_T = 3 \text{ A} \)

#### Solution:
1. Total Resistance (\( R_T \)):
The resistors are in series:
\[
R_T = R_1 + R_2 + R_3 = 8 + 12 + 6 = 26 \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law:
\[
V_T = I_T \cdot R_T = 3 \cdot 26 = 78 \text{ V}
\]

3. Voltage Drops (\( V_1, V_2, V_3 \)):
Using Ohm's Law for each resistor:
\[
V_1 = I_T \cdot R_1 = 3 \cdot 8 = 24 \text{ V}
\]
\[
V_2 = I_T \cdot R_2 = 3 \cdot 12 = 36 \text{ V}
\]
\[
V_3 = I_T \cdot R_3 = 3 \cdot 6 = 18 \text{ V}
\]

#### Final Answers:
\[
R_T = 26 \Omega, \quad I_T = 3 \text{ A}, \quad V_T = 78 \text{ V}
\]
\[
V_1 = 24 \text{ V}, \quad V_2 = 36 \text{ V}, \quad V_3 = 18 \text{ V}
\]

---

Circuit 2:

#### Given:
- Resistors: \( R_1 = 30 \Omega \), \( R_2 = 30 \Omega \), \( R_3 = 5 \Omega \)
- Voltage source: \( V_S = 40 \text{ V} \)

#### Solution:
1. Total Resistance (\( R_T \)):
\( R_1 \) and \( R_2 \) are in parallel:
\[
R_{12} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{30 \cdot 30}{30 + 30} = \frac{900}{60} = 15 \Omega
\]
\( R_{12} \) is in series with \( R_3 \):
\[
R_T = R_{12} + R_3 = 15 + 5 = 20 \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_S}{R_T} = \frac{40}{20} = 2 \text{ A}
\]

3. Voltage Drop across \( R_3 \) (\( V_3 \)):
\[
V_3 = I_T \cdot R_3 = 2 \cdot 5 = 10 \text{ V}
\]

4. Voltage Drop across \( R_{12} \) (\( V_{12} \)):
\[
V_{12} = V_S - V_3 = 40 - 10 = 30 \text{ V}
\]

5. Currents through \( R_1 \) and \( R_2 \) (\( I_1, I_2 \)):
Since \( R_1 \) and \( R_2 \) are in parallel:
\[
I_1 = \frac{V_{12}}{R_1} = \frac{30}{30} = 1 \text{ A}
\]
\[
I_2 = \frac{V_{12}}{R_2} = \frac{30}{30} = 1 \text{ A}
\]

6. Voltage Drops across \( R_1 \) and \( R_2 \) (\( V_1, V_2 \)):
\[
V_1 = I_1 \cdot R_1 = 1 \cdot 30 = 30 \text{ V}
\]
\[
V_2 = I_2 \cdot R_2 = 1 \cdot 30 = 30 \text{ V}
\]

#### Final Answers:
\[
R_T = 20 \Omega, \quad I_T = 2 \text{ A}, \quad V_T = 40 \text{ V}
\]
\[
I_1 = 1 \text{ A}, \quad I_2 = 1 \text{ A}, \quad I_3 = 2 \text{ A}
\]
\[
V_1 = 30 \text{ V}, \quad V_2 = 30 \text{ V}, \quad V_3 = 10 \text{ V}
\]

---

Circuit 3:

#### Given:
- Resistors: \( R_1 = 5 \Omega \), \( R_2 = 6 \Omega \), \( R_3 = 4 \Omega \), \( R_4 = 10 \Omega \)
- Voltage source: \( V_S = 20 \text{ V} \)

#### Solution:
1. Total Resistance (\( R_T \)):
\( R_2 \) and \( R_3 \) are in parallel:
\[
R_{23} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{6 \cdot 4}{6 + 4} = \frac{24}{10} = 2.4 \Omega
\]
\( R_1 \) and \( R_{23} \) are in series:
\[
R_{123} = R_1 + R_{23} = 5 + 2.4 = 7.4 \Omega
\]
\( R_{123} \) and \( R_4 \) are in parallel:
\[
R_T = \frac{R_{123} \cdot R_4}{R_{123} + R_4} = \frac{7.4 \cdot 10}{7.4 + 10} = \frac{74}{17.4} \approx 4.25 \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_S}{R_T} = \frac{20}{4.25} \approx 4.71 \text{ A}
\]

3. Current through \( R_4 \) (\( I_4 \)):
\[
I_4 = \frac{V_S}{R_4} = \frac{20}{10} = 2 \text{ A}
\]

4. Current through \( R_{123} \) (\( I_{123} \)):
\[
I_{123} = I_T - I_4 = 4.71 - 2 = 2.71 \text{ A}
\]

5. Voltage Drop across \( R_4 \) (\( V_4 \)):
\[
V_4 = I_4 \cdot R_4 = 2 \cdot 10 = 20 \text{ V}
\]

6. Voltage Drop across \( R_{123} \) (\( V_{123} \)):
\[
V_{123} = I_{123} \cdot R_{123} = 2.71 \cdot 7.4 \approx 20 \text{ V}
\]

7. Voltages across \( R_1 \), \( R_2 \), and \( R_3 \):
\[
V_1 = I_{123} \cdot R_1 = 2.71 \cdot 5 \approx 13.55 \text{ V}
\]
\[
V_{23} = V_{123} - V_1 = 20 - 13.55 \approx 6.45 \text{ V}
\]
\[
I_2 = \frac{V_{23}}{R_2} = \frac{6.45}{6} \approx 1.075 \text{ A}
\]
\[
I_3 = \frac{V_{23}}{R_3} = \frac{6.45}{4} \approx 1.6125 \text{ A}
\]

#### Final Answers:
\[
R_T \approx 4.25 \Omega, \quad I_T \approx 4.71 \text{ A}, \quad I_4 = 2 \text{ A}
\]
\[
V_1 \approx 13.55 \text{ V}, \quad V_4 = 20 \text{ V}
\]

---

Circuit 4:

#### Given:
- Resistors: \( R_1 = 4 \Omega \), \( R_2 = 4 \Omega \), \( R_3 = 6 \Omega \), \( R_4 = 7 \Omega \)
- Current source: \( I_S = 4 \text{ A} \)

#### Solution:
1. Total Resistance (\( R_T \)):
\( R_1 \) and \( R_2 \) are in parallel:
\[
R_{12} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{4 \cdot 4}{4 + 4} = \frac{16}{8} = 2 \Omega
\]
\( R_{12} \) is in series with \( R_3 \):
\[
R_{123} = R_{12} + R_3 = 2 + 6 = 8 \Omega
\]
\( R_{123} \) and \( R_4 \) are in parallel:
\[
R_T = \frac{R_{123} \cdot R_4}{R_{123} + R_4} = \frac{8 \cdot 7}{8 + 7} = \frac{56}{15} \approx 3.73 \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law:
\[
V_T = I_S \cdot R_T = 4 \cdot \frac{56}{15} = \frac{224}{15} \approx 14.93 \text{ V}
\]

3. Voltage Drop across \( R_4 \) (\( V_4 \)):
\[
V_4 = I_S \cdot R_4 = 4 \cdot 7 = 28 \text{ V}
\]

4. Voltage Drop across \( R_{123} \) (\( V_{123} \)):
\[
V_{123} = V_T = \frac{224}{15} \approx 14.93 \text{ V}
\]

5. Voltages across \( R_1 \) and \( R_2 \):
\[
V_{12} = V_{123} = \frac{224}{15} \approx 14.93 \text{ V}
\]
\[
I_1 = \frac{V_{12}}{R_1} = \frac{\frac{224}{15}}{4} = \frac{224}{60} \approx 3.73 \text{ A}
\]
\[
I_2 = \frac{V_{12}}{R_2} = \frac{\frac{224}{15}}{4} = \frac{224}{60} \approx 3.73 \text{ A}
\]

#### Final Answers:
\[
R_T \approx 3.73 \Omega, \quad I_T = 4 \text{ A}, \quad V_T \approx 14.93 \text{ V}
\]
\[
I_1 \approx 3.73 \text{ A}, \quad I_2 \approx 3.73 \text{ A}, \quad I_3 = 4 \text{ A}, \quad I_4 = 4 \text{ A}
\]
\[
V_1 \approx 14.93 \text{ V}, \quad V_2 \approx 14.93 \text{ V}, \quad V_3 \approx 14.93 \text{ V}, \quad V_4 = 28 \text{ V}
\]

---

Circuit 5:

#### Given:
- Resistors: \( R_1 = 2 \Omega \), \( R_2 = 2 \Omega \), \( R_3 = 2 \Omega \)
- Voltage sources: \( V_1 = 4 \text{ V} \), \( V_2 = 8 \text{ V} \)

#### Solution:
1. Equivalent Resistance (\( R_{eq} \)):
All resistors are in parallel:
\[
R_{eq} = \frac{R_1 \cdot R_2 \cdot R_3}{R_1 + R_2 + R_3} = \frac{2 \cdot 2 \cdot 2}{2 + 2 + 2} = \frac{8}{6} = \frac{4}{3} \Omega
\]

2. Total Current (\( I_T \)):
Using superposition:
- Contribution from \( V_1 \):
\[
I_{T1} = \frac{V_1}{R_{eq}} = \frac{4}{\frac{4}{3}} = 3 \text{ A}
\]
- Contribution from \( V_2 \):
\[
I_{T2} = \frac{V_2}{R_{eq}} = \frac{8}{\frac{4}{3}} = 6 \text{ A}
\]
Total current:
\[
I_T = I_{T1} + I_{T2} = 3 + 6 = 9 \text{ A}
\]

3. Current through each resistor:
Using current division:
\[
I_1 = I_2 = I_3 = \frac{I_T}{3} = \frac{9}{3} = 3 \text{ A}
\]

4. Voltage across each resistor:
Using Ohm's Law:
\[
V_1 = I_1 \cdot R_1 = 3 \cdot 2 = 6 \text{ V}
\]
\[
V_2 = I_2 \cdot R_2 = 3 \cdot 2 = 6 \text{ V}
\]
\[
V_3 = I_3 \cdot R_3 = 3 \cdot 2 = 6 \text{ V}
\]

#### Final Answers:
\[
R_2 = 2 \Omega, \quad R_T = \frac{4}{3} \Omega, \quad V_T = 6 \text{ V}
\]
\[
I_1 = 3 \text{ A}, \quad V_1 = 6 \text{ V}
\]

---

Circuit 6:

#### Given:
- Resistors: \( R_1 = 6 \Omega \), \( R_2 = 6 \Omega \), \( R_3 = 12 \Omega \)
- Voltage source: \( V_S = 30 \text{ V} \)

#### Solution:
1. Total Resistance (\( R_T \)):
\( R_1 \) and \( R_2 \) are in parallel:
\[
R_{12} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{6 \cdot 6}{6 + 6} = \frac{36}{12} = 3 \Omega
\]
\( R_{12} \) is in series with \( R_3 \):
\[
R_T = R_{12} + R_3 = 3 + 12 = 15 \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_S}{R_T} = \frac{30}{15} = 2 \text{ A}
\]

3. Voltage Drop across \( R_3 \) (\( V_3 \)):
\[
V_3 = I_T \cdot R_3 = 2 \cdot 12 = 24 \text{ V}
\]

4. Voltage Drop across \( R_{12} \) (\( V_{12} \)):
\[
V_{12} = V_S - V_3 = 30 - 24 = 6 \text{ V}
\]

5. Currents through \( R_1 \) and \( R_2 \) (\( I_1, I_2 \)):
Since \( R_1 \) and \( R_2 \) are in parallel:
\[
I_1 = \frac{V_{12}}{R_1} = \frac{6}{6} = 1 \text{ A}
\]
\[
I_2 = \frac{V_{12}}{R_2} = \frac{6}{6} = 1 \text{ A}
\]

6. Voltage Drops across \( R_1 \) and \( R_2 \) (\( V_1, V_2 \)):
\[
V_1 = I_1 \cdot R_1 = 1 \cdot 6 = 6 \text{ V}
\]
\[
V_2 = I_2 \cdot R_2 = 1 \cdot 6 = 6 \text{ V}
\]

#### Final Answers:
\[
R_T = 15 \Omega, \quad I_T = 2 \text{ A}, \quad V_T = 30 \text{ V}
\]
\[
I_1 = 1 \text{ A}, \quad I_2 = 1 \text{ A}, \quad I_3 = 2 \text{ A}
\]
\[
V_1 = 6 \text{ V}, \quad V_2 = 6 \text{ V}, \quad V_3 = 24 \text{ V}
\]

---

Final Answer Summary:

\[
\boxed{
\begin{array}{l}
\text{Circuit 1: } R_T = 26 \Omega, I_T = 3 \text{ A}, V_T = 78 \text{ V}, V_1 = 24 \text{ V}, V_2 = 36 \text{ V}, V_3 = 18 \text{ V} \\
\text{Circuit 2: } R_T = 20 \Omega, I_T = 2 \text{ A}, V_T = 40 \text{ V}, I_1 = 1 \text{ A}, I_2 = 1 \text{ A}, I_3 = 2 \text{ A}, V_1 = 30 \text{ V}, V_2 = 30 \text{ V}, V_3 = 10 \text{ V} \\
\text{Circuit 3: } R_T \approx 4.25 \Omega, I_T \approx 4.71 \text{ A}, I_4 = 2 \text{ A}, V_1 \approx 13.55 \text{ V}, V_4 = 20 \text{ V} \\
\text{Circuit 4: } R_T \approx 3.73 \Omega, I_T = 4 \text{ A}, V_T \approx 14.93 \text{ V}, I_1 \approx 3.73 \text{ A}, I_2 \approx 3.73 \text{ A}, I_3 = 4 \text{ A}, I_4 = 4 \text{ A}, V_1 \approx 14.93 \text{ V}, V_2 \approx 14.93 \text{ V}, V_3 \approx 14.93 \text{ V}, V_4 = 28 \text{ V} \\
\text{Circuit 5: } R_2 = 2 \Omega, R_T = \frac{4}{3} \Omega, V_T = 6 \text{ V}, I_1 = 3 \text{ A}, V_1 = 6 \text{ V} \\
\text{Circuit 6: } R_T = 15 \Omega, I_T = 2 \text{ A}, V_T = 30 \text{ V}, I_1 = 1 \text{ A}, I_2 = 1 \text{ A}, I_3 = 2 \text{ A}, V_1 = 6 \text{ V}, V_2 = 6 \text{ V}, V_3 = 24 \text{ V}
\end{array}
}
\]