Series-Parallel Circuit Problems Worksheet with Diagrams and Calculations

Problem Analysis:

The worksheet involves solving series-parallel circuits using Ohm's Law and the rules for combining resistances in series and parallel. Let's solve each circuit step by step.

---

Circuit 1:

#### Circuit Diagram:
- Voltage source: \( V = 24 \, \text{V} \)
- Resistors:
- Top branch: \( R_1 = 100 \, \Omega \), \( R_2 = 2.30 \, \Omega \)
- Bottom branch: \( R_3 = 330 \, \Omega \), \( R_4 = 200 \, \Omega \)

#### Steps:
1. Calculate Equivalent Resistance (\( R_{\text{eq}} \)):
- The top branch has \( R_1 \) and \( R_2 \) in series:
\[
R_{\text{top}} = R_1 + R_2 = 100 \, \Omega + 2.30 \, \Omega = 102.30 \, \Omega
\]
- The bottom branch has \( R_3 \) and \( R_4 \) in series:
\[
R_{\text{bottom}} = R_3 + R_4 = 330 \, \Omega + 200 \, \Omega = 530 \, \Omega
\]
- The two branches are in parallel:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{top}}} + \frac{1}{R_{\text{bottom}}}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{102.30} + \frac{1}{530}
\]
\[
\frac{1}{R_{\text{eq}}} = 0.00977 + 0.00189 = 0.01166
\]
\[
R_{\text{eq}} = \frac{1}{0.01166} \approx 85.8 \, \Omega
\]

2. Calculate Total Current (\( I_{\text{total}} \)):
Using Ohm's Law:
\[
I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{24 \, \text{V}}{85.8 \, \Omega} \approx 0.28 \, \text{A}
\]

3. Distribute Currents:
- Current through the top branch:
\[
I_{\text{top}} = \frac{V}{R_{\text{top}}} = \frac{24 \, \text{V}}{102.30 \, \Omega} \approx 0.235 \, \text{A}
\]
- Current through the bottom branch:
\[
I_{\text{bottom}} = \frac{V}{R_{\text{bottom}}} = \frac{24 \, \text{V}}{530 \, \Omega} \approx 0.045 \, \text{A}
\]

4. Calculate Voltages Across Each Resistor:
- Top branch:
\[
V_1 = I_{\text{top}} \cdot R_1 = 0.235 \, \text{A} \cdot 100 \, \Omega = 23.5 \, \text{V}
\]
\[
V_2 = I_{\text{top}} \cdot R_2 = 0.235 \, \text{A} \cdot 2.30 \, \Omega = 0.54 \, \text{V}
\]
- Bottom branch:
\[
V_3 = I_{\text{bottom}} \cdot R_3 = 0.045 \, \text{A} \cdot 330 \, \Omega = 14.85 \, \text{V}
\]
\[
V_4 = I_{\text{bottom}} \cdot R_4 = 0.045 \, \text{A} \cdot 200 \, \Omega = 9.0 \, \text{V}
\]

#### Final Answers for Circuit 1:
\[
\boxed{
\begin{aligned}
R_{\text{eq}} &= 85.8 \, \Omega \\
I_1 &= 0.235 \, \text{A}, & I_2 &= 0.235 \, \text{A}, & I_3 &= 0.045 \, \text{A}, & I_4 &= 0.045 \, \text{A} \\
V_1 &= 23.5 \, \text{V}, & V_2 &= 0.54 \, \text{V}, & V_3 &= 14.85 \, \text{V}, & V_4 &= 9.0 \, \text{V}
\end{aligned}
}
\]

---

Circuit 2:

#### Circuit Diagram:
- Voltage source: \( V = 220 \, \text{V} \)
- Resistors:
- Left branch: \( R_1 = 50 \, \Omega \), \( R_5 = 20 \, \Omega \)
- Middle branch: \( R_2 = 60 \, \Omega \), \( R_3 = 20 \, \Omega \)
- Right branch: \( R_4 = 70 \, \Omega \), \( R_5 = 20 \, \Omega \)

#### Steps:
1. Calculate Equivalent Resistance (\( R_{\text{eq}} \)):
- Each branch is a series combination:
\[
R_{\text{left}} = R_1 + R_5 = 50 \, \Omega + 20 \, \Omega = 70 \, \Omega
\]
\[
R_{\text{middle}} = R_2 + R_3 = 60 \, \Omega + 20 \, \Omega = 80 \, \Omega
\]
\[
R_{\text{right}} = R_4 + R_5 = 70 \, \Omega + 20 \, \Omega = 90 \, \Omega
\]
- The three branches are in parallel:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{left}}} + \frac{1}{R_{\text{middle}}} + \frac{1}{R_{\text{right}}}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{70} + \frac{1}{80} + \frac{1}{90}
\]
\[
\frac{1}{R_{\text{eq}}} = 0.0143 + 0.0125 + 0.0111 = 0.038
\]
\[
R_{\text{eq}} = \frac{1}{0.038} \approx 26.3 \, \Omega
\]

2. Calculate Total Current (\( I_{\text{total}} \)):
Using Ohm's Law:
\[
I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{220 \, \text{V}}{26.3 \, \Omega} \approx 8.37 \, \text{A}
\]

3. Distribute Currents:
- Current through the left branch:
\[
I_{\text{left}} = \frac{V}{R_{\text{left}}} = \frac{220 \, \text{V}}{70 \, \Omega} \approx 3.14 \, \text{A}
\]
- Current through the middle branch:
\[
I_{\text{middle}} = \frac{V}{R_{\text{middle}}} = \frac{220 \, \text{V}}{80 \, \Omega} \approx 2.75 \, \text{A}
\]
- Current through the right branch:
\[
I_{\text{right}} = \frac{V}{R_{\text{right}}} = \frac{220 \, \text{V}}{90 \, \Omega} \approx 2.44 \, \text{A}
\]

4. Calculate Voltages Across Each Resistor:
- Left branch:
\[
V_1 = I_{\text{left}} \cdot R_1 = 3.14 \, \text{A} \cdot 50 \, \Omega = 157 \, \text{V}
\]
\[
V_5 = I_{\text{left}} \cdot R_5 = 3.14 \, \text{A} \cdot 20 \, \Omega = 62.8 \, \text{V}
\]
- Middle branch:
\[
V_2 = I_{\text{middle}} \cdot R_2 = 2.75 \, \text{A} \cdot 60 \, \Omega = 165 \, \text{V}
\]
\[
V_3 = I_{\text{middle}} \cdot R_3 = 2.75 \, \text{A} \cdot 20 \, \Omega = 55 \, \text{V}
\]
- Right branch:
\[
V_4 = I_{\text{right}} \cdot R_4 = 2.44 \, \text{A} \cdot 70 \, \Omega = 170.8 \, \text{V}
\]
\[
V_5 = I_{\text{right}} \cdot R_5 = 2.44 \, \text{A} \cdot 20 \, \Omega = 48.8 \, \text{V}
\]

#### Final Answers for Circuit 2:
\[
\boxed{
\begin{aligned}
R_{\text{eq}} &= 26.3 \, \Omega \\
I_1 &= 3.14 \, \text{A}, & I_2 &= 2.75 \, \text{A}, & I_3 &= 2.44 \, \text{A}, & I_4 &= 2.44 \, \text{A}, & I_5 &= 3.14 \, \text{A}, & I_6 &= 2.75 \, \text{A} \\
V_1 &= 157 \, \text{V}, & V_2 &= 165 \, \text{V}, & V_3 &= 55 \, \text{V}, & V_4 &= 170.8 \, \text{V}, & V_5 &= 62.8 \, \text{V}, & V_6 &= 48.8 \, \text{V}
\end{aligned}
}
\]

---

Circuit 3:

#### Circuit Diagram:
- Voltage source: \( V = 120 \, \text{V} \)
- Resistors:
- Left branch: \( R_2 = 200 \, \Omega \), \( R_3 = 200 \, \Omega \)
- Right branch: \( R_2 = 200 \, \Omega \), \( R_3 = 200 \, \Omega \)

#### Steps:
1. Calculate Equivalent Resistance (\( R_{\text{eq}} \)):
- Each branch is a series combination:
\[
R_{\text{left}} = R_2 + R_3 = 200 \, \Omega + 200 \, \Omega = 400 \, \Omega
\]
\[
R_{\text{right}} = R_2 + R_3 = 200 \, \Omega + 200 \, \Omega = 400 \, \Omega
\]
- The two branches are in parallel:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{left}}} + \frac{1}{R_{\text{right}}}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{400} + \frac{1}{400} = \frac{2}{400} = \frac{1}{200}
\]
\[
R_{\text{eq}} = 200 \, \Omega
\]

2. Calculate Total Current (\( I_{\text{total}} \)):
Using Ohm's Law:
\[
I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{120 \, \text{V}}{200 \, \Omega} = 0.6 \, \text{A}
\]

3. Distribute Currents:
- Current through each branch (since they are identical):
\[
I_{\text{left}} = I_{\text{right}} = \frac{I_{\text{total}}}{2} = \frac{0.6 \, \text{A}}{2} = 0.3 \, \text{A}
\]

4. Calculate Voltages Across Each Resistor:
- Left branch:
\[
V_1 = I_{\text{left}} \cdot R_2 = 0.3 \, \text{A} \cdot 200 \, \Omega = 60 \, \text{V}
\]
\[
V_2 = I_{\text{left}} \cdot R_3 = 0.3 \, \text{A} \cdot 200 \, \Omega = 60 \, \text{V}
\]
- Right branch:
\[
V_3 = I_{\text{right}} \cdot R_2 = 0.3 \, \text{A} \cdot 200 \, \Omega = 60 \, \text{V}
\]
\[
V_4 = I_{\text{right}} \cdot R_3 = 0.3 \, \text{A} \cdot 200 \, \Omega = 60 \, \text{V}
\]

#### Final Answers for Circuit 3:
\[
\boxed{
\begin{aligned}
R_{\text{eq}} &= 200 \, \Omega \\
I_1 &= 0.3 \, \text{A}, & I_2 &= 0.3 \, \text{A}, & I_3 &= 0.3 \, \text{A}, & I_4 &= 0.3 \, \text{A} \\
V_1 &= 60 \, \text{V}, & V_2 &= 60 \, \text{V}, & V_3 &= 60 \, \text{V}, & V_4 &= 60 \, \text{V}
\end{aligned}
}
\]

---

Final Summary of All Circuits:

1. Circuit 1:
\[
\boxed{
\begin{aligned}
R_{\text{eq}} &= 85.8 \, \Omega \\
I_1 &= 0.235 \, \text{A}, & I_2 &= 0.235 \, \text{A}, & I_3 &= 0.045 \, \text{A}, & I_4 &= 0.045 \, \text{A} \\
V_1 &= 23.5 \, \text{V}, & V_2 &= 0.54 \, \text{V}, & V_3 &= 14.85 \, \text{V}, & V_4 &= 9.0 \, \text{V}
\end{aligned}
}
\]

2. Circuit 2:
\[
\boxed{
\begin{aligned}
R_{\text{eq}} &= 26.3 \, \Omega \\
I_1 &= 3.14 \, \text{A}, & I_2 &= 2.75 \, \text{A}, & I_3 &= 2.44 \, \text{A}, & I_4 &= 2.44 \, \text{A}, & I_5 &= 3.14 \, \text{A}, & I_6 &= 2.75 \, \text{A} \\
V_1 &= 157 \, \text{V}, & V_2 &= 165 \, \text{V}, & V_3 &= 55 \, \text{V}, & V_4 &= 170.8 \, \text{V}, & V_5 &= 62.8 \, \text{V}, & V_6 &= 48.8 \, \text{V}
\end{aligned}
}
\]

3. Circuit 3:
\[
\boxed{
\begin{aligned}
R_{\text{eq}} &= 200 \, \Omega \\
I_1 &= 0.3 \, \text{A}, & I_2 &= 0.3 \, \text{A}, & I_3 &= 0.3 \, \text{A}, & I_4 &= 0.3 \, \text{A} \\
V_1 &= 60 \, \text{V}, & V_2 &= 60 \, \text{V}, & V_3 &= 60 \, \text{V}, & V_4 &= 60 \, \text{V}
\end{aligned}
}
\]