Worksheet with six series circuit problems, each featuring a battery, resistors, and handwritten calculations for voltage, current, and resistance.

Problem Analysis and Solution

The worksheet involves solving problems related to series circuits. In a series circuit:
1. The current is the same through every component.
2. The total voltage supplied by the battery is equal to the sum of the voltage drops across each resistor.
3. The total resistance is the sum of the individual resistances.

We will solve each problem step by step using these principles.

---

#### Problem 1:
- Circuit Diagram: A series circuit with two resistors \( R_1 = 10 \, \Omega \) and \( R_2 = 20 \, \Omega \), powered by a 90 V battery.
- Given:
- \( R_1 = 10 \, \Omega \)
- \( R_2 = 20 \, \Omega \)
- Battery voltage \( V = 90 \, \text{V} \)

Solution:
1. Calculate Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 10 \, \Omega + 20 \, \Omega = 30 \, \Omega
\]

2. Calculate Total Current (\( I_T \)):
Using Ohm's Law: \( V = I \cdot R \)
\[
I_T = \frac{V}{R_T} = \frac{90 \, \text{V}}{30 \, \Omega} = 3 \, \text{A}
\]

3. Calculate Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I_T \cdot R_1 = 3 \, \text{A} \cdot 10 \, \Omega = 30 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I_T \cdot R_2 = 3 \, \text{A} \cdot 20 \, \Omega = 60 \, \text{V}
\]

4. Verify Total Voltage:
\[
V_1 + V_2 = 30 \, \text{V} + 60 \, \text{V} = 90 \, \text{V} \quad \text{(Correct)}
\]

Final Answers for Problem 1:
\[
R_T = 30 \, \Omega, \quad I_T = 3 \, \text{A}, \quad V_1 = 30 \, \text{V}, \quad V_2 = 60 \, \text{V}
\]

---

#### Problem 2:
- Circuit Diagram: A series circuit with three resistors \( R_1 = 6 \, \Omega \), \( R_2 = 14 \, \Omega \), and \( R_3 = 10 \, \Omega \), powered by a 60 V battery.
- Given:
- \( R_1 = 6 \, \Omega \)
- \( R_2 = 14 \, \Omega \)
- \( R_3 = 10 \, \Omega \)
- Battery voltage \( V = 60 \, \text{V} \)

Solution:
1. Calculate Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 = 6 \, \Omega + 14 \, \Omega + 10 \, \Omega = 30 \, \Omega
\]

2. Calculate Total Current (\( I_T \)):
Using Ohm's Law: \( V = I \cdot R \)
\[
I_T = \frac{V}{R_T} = \frac{60 \, \text{V}}{30 \, \Omega} = 2 \, \text{A}
\]

3. Calculate Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I_T \cdot R_1 = 2 \, \text{A} \cdot 6 \, \Omega = 12 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I_T \cdot R_2 = 2 \, \text{A} \cdot 14 \, \Omega = 28 \, \text{V}
\]
- Across \( R_3 \):
\[
V_3 = I_T \cdot R_3 = 2 \, \text{A} \cdot 10 \, \Omega = 20 \, \text{V}
\]

4. Verify Total Voltage:
\[
V_1 + V_2 + V_3 = 12 \, \text{V} + 28 \, \text{V} + 20 \, \text{V} = 60 \, \text{V} \quad \text{(Correct)}
\]

Final Answers for Problem 2:
\[
R_T = 30 \, \Omega, \quad I_T = 2 \, \text{A}, \quad V_1 = 12 \, \text{V}, \quad V_2 = 28 \, \text{V}, \quad V_3 = 20 \, \text{V}
\]

---

#### Problem 3:
- Circuit Diagram: A series circuit with two resistors \( R_1 = 10 \, \Omega \) and \( R_2 \), powered by a 75 V battery. The current is given as 5 A, and the voltage drop across \( R_2 \) is 25 V.
- Given:
- \( R_1 = 10 \, \Omega \)
- Battery voltage \( V = 75 \, \text{V} \)
- Current \( I_T = 5 \, \text{A} \)
- Voltage drop across \( R_2 \): \( V_2 = 25 \, \text{V} \)

Solution:
1. Calculate Voltage Drop Across \( R_1 \):
Using Ohm's Law: \( V = I \cdot R \)
\[
V_1 = I_T \cdot R_1 = 5 \, \text{A} \cdot 10 \, \Omega = 50 \, \text{V}
\]

2. Calculate Resistance \( R_2 \):
Using Ohm's Law: \( R = \frac{V}{I} \)
\[
R_2 = \frac{V_2}{I_T} = \frac{25 \, \text{V}}{5 \, \text{A}} = 5 \, \Omega
\]

3. Verify Total Voltage:
\[
V_1 + V_2 = 50 \, \text{V} + 25 \, \text{V} = 75 \, \text{V} \quad \text{(Correct)}
\]

Final Answers for Problem 3:
\[
V_1 = 50 \, \text{V}, \quad I_2 = 5 \, \text{A}, \quad R_2 = 5 \, \Omega
\]

---

#### Problem 4:
- Circuit Diagram: A series circuit with two resistors \( R_1 = 5 \, \Omega \) and \( R_2 = 15 \, \Omega \), powered by a 100 V battery. The current is given as 5 A.
- Given:
- \( R_1 = 5 \, \Omega \)
- \( R_2 = 15 \, \Omega \)
- Battery voltage \( V_T = 100 \, \text{V} \)
- Current \( I_T = 5 \, \text{A} \)

Solution:
1. Calculate Voltage Drop Across \( R_1 \):
Using Ohm's Law: \( V = I \cdot R \)
\[
V_1 = I_T \cdot R_1 = 5 \, \text{A} \cdot 5 \, \Omega = 25 \, \text{V}
\]

2. Calculate Voltage Drop Across \( R_2 \):
Using Ohm's Law: \( V = I \cdot R \)
\[
V_2 = I_T \cdot R_2 = 5 \, \text{A} \cdot 15 \, \Omega = 75 \, \text{V}
\]

3. Verify Total Voltage:
\[
V_1 + V_2 = 25 \, \text{V} + 75 \, \text{V} = 100 \, \text{V} \quad \text{(Correct)}
\]

Final Answers for Problem 4:
\[
V_1 = 25 \, \text{V}, \quad V_2 = 75 \, \text{V}, \quad V_T = 100 \, \text{V}
\]

---

#### Problem 5:
- Circuit Diagram: A series circuit with three resistors \( R_1 = 5 \, \Omega \), \( R_2 = 3 \, \Omega \), and \( R_3 = 2 \, \Omega \), powered by a 20 V battery.
- Given:
- \( R_1 = 5 \, \Omega \)
- \( R_2 = 3 \, \Omega \)
- \( R_3 = 2 \, \Omega \)
- Battery voltage \( V = 20 \, \text{V} \)

Solution:
1. Calculate Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 = 5 \, \Omega + 3 \, \Omega + 2 \, \Omega = 10 \, \Omega
\]

2. Calculate Total Current (\( I_T \)):
Using Ohm's Law: \( V = I \cdot R \)
\[
I_T = \frac{V}{R_T} = \frac{20 \, \text{V}}{10 \, \Omega} = 2 \, \text{A}
\]

3. Calculate Voltage Drops:
- Across \( R_1 \):
\[
V_1 = I_T \cdot R_1 = 2 \, \text{A} \cdot 5 \, \Omega = 10 \, \text{V}
\]
- Across \( R_2 \):
\[
V_2 = I_T \cdot R_2 = 2 \, \text{A} \cdot 3 \, \Omega = 6 \, \text{V}
\]
- Across \( R_3 \):
\[
V_3 = I_T \cdot R_3 = 2 \, \text{A} \cdot 2 \, \Omega = 4 \, \text{V}
\]

4. Verify Total Voltage:
\[
V_1 + V_2 + V_3 = 10 \, \text{V} + 6 \, \text{V} + 4 \, \text{V} = 20 \, \text{V} \quad \text{(Correct)}
\]

Final Answers for Problem 5:
\[
R_T = 10 \, \Omega, \quad I_T = 2 \, \text{A}, \quad V_1 = 10 \, \text{V}, \quad V_2 = 6 \, \text{V}, \quad V_3 = 4 \, \text{V}
\]

---

#### Problem 6:
- Circuit Diagram: A series circuit with three resistors \( R_1 \), \( R_2 \), and \( R_3 = 10 \, \Omega \), powered by a 40 V battery. The voltage drop across \( R_3 \) is 10 V, and the voltage drop across \( R_2 \) is 20 V.
- Given:
- \( R_3 = 10 \, \Omega \)
- Battery voltage \( V = 40 \, \text{V} \)
- Voltage drop across \( R_3 \): \( V_3 = 10 \, \text{V} \)
- Voltage drop across \( R_2 \): \( V_2 = 20 \, \text{V} \)

Solution:
1. Calculate Voltage Drop Across \( R_1 \):
\[
V_1 = V - V_2 - V_3 = 40 \, \text{V} - 20 \, \text{V} - 10 \, \text{V} = 10 \, \text{V}
\]

2. Calculate Total Current (\( I_T \)):
Using the voltage drop across \( R_3 \):
\[
I_T = \frac{V_3}{R_3} = \frac{10 \, \text{V}}{10 \, \Omega} = 1 \, \text{A}
\]

3. Calculate Resistances \( R_1 \) and \( R_2 \):
- For \( R_1 \):
\[
R_1 = \frac{V_1}{I_T} = \frac{10 \, \text{V}}{1 \, \text{A}} = 10 \, \Omega
\]
- For \( R_2 \):
\[
R_2 = \frac{V_2}{I_T} = \frac{20 \, \text{V}}{1 \, \text{A}} = 20 \, \Omega
\]

4. Verify Total Voltage:
\[
V_1 + V_2 + V_3 = 10 \, \text{V} + 20 \, \text{V} + 10 \, \text{V} = 40 \, \text{V} \quad \text{(Correct)}
\]

Final Answers for Problem 6:
\[
I_3 = 1 \, \text{A}, \quad I_1 = 1 \, \text{A}, \quad V_1 = 10 \, \text{V}, \quad R_1 = 10 \, \Omega, \quad R_2 = 20 \, \Omega
\]

---

Final Answers Summary:

1. \( R_T = 30 \, \Omega, \, I_T = 3 \, \text{A}, \, V_1 = 30 \, \text{V}, \, V_2 = 60 \, \text{V} \)
2. \( R_T = 30 \, \Omega, \, I_T = 2 \, \text{A}, \, V_1 = 12 \, \text{V}, \, V_2 = 28 \, \text{V}, \, V_3 = 20 \, \text{V} \)
3. \( V_1 = 50 \, \text{V}, \, I_2 = 5 \, \text{A}, \, R_2 = 5 \, \Omega \)
4. \( V_1 = 25 \, \text{V}, \, V_2 = 75 \, \text{V}, \, V_T = 100 \, \text{V} \)
5. \( R_T = 10 \, \Omega, \, I_T = 2 \, \text{A}, \, V_1 = 10 \, \text{V}, \, V_2 = 6 \, \text{V}, \, V_3 = 4 \, \text{V} \)
6. \( I_3 = 1 \, \text{A}, \, I_1 = 1 \, \text{A}, \, V_1 = 10 \, \text{V}, \, R_1 = 10 \, \Omega, \, R_2 = 20 \, \Omega \)

\boxed{\text{See detailed solutions above}}