Series Circuit Problems Worksheet - Practice calculating voltage, current, and resistance in series circuits.

To solve the problems in the worksheet, we need to analyze each circuit and apply the principles of Ohm's Law and Kirchhoff's Laws. Let's go through each circuit step by step.

---

Circuit 1:

#### Given:
- \( V_s = 12 \, \text{V} \)
- \( R_1 = 4 \, \Omega \)
- \( R_2 = 6 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_{\text{total}} \)):
Since \( R_1 \) and \( R_2 \) are in series:
\[
R_{\text{total}} = R_1 + R_2 = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega
\]

2. Total Current (\( I \)):
Using Ohm's Law:
\[
I = \frac{V_s}{R_{\text{total}}} = \frac{12 \, \text{V}}{10 \, \Omega} = 1.2 \, \text{A}
\]

3. Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I \cdot R_1 = 1.2 \, \text{A} \cdot 4 \, \Omega = 4.8 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I \cdot R_2 = 1.2 \, \text{A} \cdot 6 \, \Omega = 7.2 \, \text{V}
\]

4. Verification:
The sum of voltage drops should equal the source voltage:
\[
V_1 + V_2 = 4.8 \, \text{V} + 7.2 \, \text{V} = 12 \, \text{V} \quad \text{(Correct)}
\]

#### Answers:
\[
I = 1.2 \, \text{A}, \quad V_1 = 4.8 \, \text{V}, \quad V_2 = 7.2 \, \text{V}
\]

---

Circuit 2:

#### Given:
- \( V_s = 15 \, \text{V} \)
- \( R_1 = 5 \, \Omega \)
- \( R_2 = 10 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_{\text{total}} \)):
Since \( R_1 \) and \( R_2 \) are in series:
\[
R_{\text{total}} = R_1 + R_2 = 5 \, \Omega + 10 \, \Omega = 15 \, \Omega
\]

2. Total Current (\( I \)):
Using Ohm's Law:
\[
I = \frac{V_s}{R_{\text{total}}} = \frac{15 \, \text{V}}{15 \, \Omega} = 1 \, \text{A}
\]

3. Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I \cdot R_1 = 1 \, \text{A} \cdot 5 \, \Omega = 5 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I \cdot R_2 = 1 \, \text{A} \cdot 10 \, \Omega = 10 \, \text{V}
\]

4. Verification:
The sum of voltage drops should equal the source voltage:
\[
V_1 + V_2 = 5 \, \text{V} + 10 \, \text{V} = 15 \, \text{V} \quad \text{(Correct)}
\]

#### Answers:
\[
I = 1 \, \text{A}, \quad V_1 = 5 \, \text{V}, \quad V_2 = 10 \, \text{V}
\]

---

Circuit 3:

#### Given:
- \( V_s = 9 \, \text{V} \)
- \( R_1 = 3 \, \Omega \)
- \( R_2 = 6 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_{\text{total}} \)):
Since \( R_1 \) and \( R_2 \) are in series:
\[
R_{\text{total}} = R_1 + R_2 = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega
\]

2. Total Current (\( I \)):
Using Ohm's Law:
\[
I = \frac{V_s}{R_{\text{total}}} = \frac{9 \, \text{V}}{9 \, \Omega} = 1 \, \text{A}
\]

3. Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I \cdot R_1 = 1 \, \text{A} \cdot 3 \, \Omega = 3 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I \cdot R_2 = 1 \, \text{A} \cdot 6 \, \Omega = 6 \, \text{V}
\]

4. Verification:
The sum of voltage drops should equal the source voltage:
\[
V_1 + V_2 = 3 \, \text{V} + 6 \, \text{V} = 9 \, \text{V} \quad \text{(Correct)}
\]

#### Answers:
\[
I = 1 \, \text{A}, \quad V_1 = 3 \, \text{V}, \quad V_2 = 6 \, \text{V}
\]

---

Circuit 4:

#### Given:
- \( V_s = 24 \, \text{V} \)
- \( R_1 = 8 \, \Omega \)
- \( R_2 = 12 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_{\text{total}} \)):
Since \( R_1 \) and \( R_2 \) are in series:
\[
R_{\text{total}} = R_1 + R_2 = 8 \, \Omega + 12 \, \Omega = 20 \, \Omega
\]

2. Total Current (\( I \)):
Using Ohm's Law:
\[
I = \frac{V_s}{R_{\text{total}}} = \frac{24 \, \text{V}}{20 \, \Omega} = 1.2 \, \text{A}
\]

3. Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I \cdot R_1 = 1.2 \, \text{A} \cdot 8 \, \Omega = 9.6 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I \cdot R_2 = 1.2 \, \text{A} \cdot 12 \, \Omega = 14.4 \, \text{V}
\]

4. Verification:
The sum of voltage drops should equal the source voltage:
\[
V_1 + V_2 = 9.6 \, \text{V} + 14.4 \, \text{V} = 24 \, \text{V} \quad \text{(Correct)}
\]

#### Answers:
\[
I = 1.2 \, \text{A}, \quad V_1 = 9.6 \, \text{V}, \quad V_2 = 14.4 \, \text{V}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{Circuit 1: } I = 1.2 \, \text{A}, \, V_1 = 4.8 \, \text{V}, \, V_2 = 7.2 \, \text{V} \\
&\text{Circuit 2: } I = 1 \, \text{A}, \, V_1 = 5 \, \text{V}, \, V_2 = 10 \, \text{V} \\
&\text{Circuit 3: } I = 1 \, \text{A}, \, V_1 = 3 \, \text{V}, \, V_2 = 6 \, \text{V} \\
&\text{Circuit 4: } I = 1.2 \, \text{A}, \, V_1 = 9.6 \, \text{V}, \, V_2 = 14.4 \, \text{V}
\end{aligned}
}
\]