Solved SERIES CIRCUIT PROBLEMS Use Ohms Law and the rules | Chegg.com - Free Printable

To solve the problems in the image, we will analyze each circuit step by step using Ohm's Law (\( V = IR \)) and Kirchhoff's Laws. Let's go through each problem systematically.

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Problem 7

Circuit Description:
- Voltage source \( V = 100 \, \text{V} \)
- Resistors \( R_1 = 20 \, \Omega \) and \( R_2 = 60 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 20 \, \Omega + 60 \, \Omega = 80 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V}{R_T} = \frac{100 \, \text{V}}{80 \, \Omega} = 1.25 \, \text{A}
\]

3. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I_T \cdot R_1 = 1.25 \, \text{A} \cdot 20 \, \Omega = 25 \, \text{V}
\]

4. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = I_T \cdot R_2 = 1.25 \, \text{A} \cdot 60 \, \Omega = 75 \, \text{V}
\]

Final Answers for Problem 7:
\[
\boxed{V_1 = 25 \, \text{V}, \, V_2 = 75 \, \text{V}, \, I_T = 1.25 \, \text{A}}
\]

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Problem 8

Circuit Description:
- Current source \( I = 2 \, \text{A} \)
- Resistors \( R_1 = 20 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 30 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 = 20 \, \Omega + 60 \, \Omega + 30 \, \Omega = 110 \, \Omega
\]

2. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I \cdot R_1 = 2 \, \text{A} \cdot 20 \, \Omega = 40 \, \text{V}
\]

3. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = I \cdot R_2 = 2 \, \text{A} \cdot 60 \, \Omega = 120 \, \text{V}
\]

4. Voltage across \( R_3 \) (\( V_3 \)):
\[
V_3 = I \cdot R_3 = 2 \, \text{A} \cdot 30 \, \Omega = 60 \, \text{V}
\]

Final Answers for Problem 8:
\[
\boxed{V_1 = 40 \, \text{V}, \, V_2 = 120 \, \text{V}, \, V_3 = 60 \, \text{V}, \, R_T = 110 \, \Omega}
\]

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Problem 9

Circuit Description:
- Current source \( I = 4 \, \text{A} \)
- Resistors \( R_1 = 10 \, \Omega \), \( R_2 \) (unknown)
- Voltage across \( R_2 \) is given as \( V_2 = 40 \, \text{V} \)

#### Solution:
1. Resistance \( R_2 \):
Using Ohm's Law:
\[
R_2 = \frac{V_2}{I} = \frac{40 \, \text{V}}{4 \, \text{A}} = 10 \, \Omega
\]

2. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I \cdot R_1 = 4 \, \text{A} \cdot 10 \, \Omega = 40 \, \text{V}
\]

3. Total Voltage (\( V_T \)):
\[
V_T = V_1 + V_2 = 40 \, \text{V} + 40 \, \text{V} = 80 \, \text{V}
\]

Final Answers for Problem 9:
\[
\boxed{R_2 = 10 \, \Omega, \, V_1 = 40 \, \text{V}, \, V_T = 80 \, \text{V}}
\]

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Problem 10

Circuit Description:
- Current source \( I = 3 \, \text{A} \)
- Resistors \( R_1 = 10 \, \Omega \), \( R_2 \) (unknown), \( R_3 = 20 \, \Omega \)
- Voltage across \( R_3 \) is given as \( V_3 = 60 \, \text{V} \)

#### Solution:
1. Resistance \( R_2 \):
Using Ohm's Law for \( R_3 \):
\[
R_3 = \frac{V_3}{I} = \frac{60 \, \text{V}}{3 \, \text{A}} = 20 \, \Omega
\]
(This confirms the given value of \( R_3 \).)

2. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I \cdot R_1 = 3 \, \text{A} \cdot 10 \, \Omega = 30 \, \text{V}
\]

3. Total Voltage (\( V_T \)):
\[
V_T = V_1 + V_3 = 30 \, \text{V} + 60 \, \text{V} = 90 \, \text{V}
\]

Final Answers for Problem 10:
\[
\boxed{V_1 = 30 \, \text{V}, \, V_T = 90 \, \text{V}, \, R_2 = 20 \, \Omega}
\]

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Problem 11

Circuit Description:
- Voltage source \( V_T = 60 \, \text{V} \)
- Resistors \( R_1 = 8 \, \Omega \), \( R_2 = 12 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 8 \, \Omega + 12 \, \Omega = 20 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{60 \, \text{V}}{20 \, \Omega} = 3 \, \text{A}
\]

3. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I_T \cdot R_1 = 3 \, \text{A} \cdot 8 \, \Omega = 24 \, \text{V}
\]

Final Answers for Problem 11:
\[
\boxed{I_T = 3 \, \text{A}, \, V_1 = 24 \, \text{V}}
\]

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Problem 12

Circuit Description:
- Voltage source \( V_T = 30 \, \text{V} \)
- Resistors \( R_1 = 6 \, \Omega \), \( R_2 = 4 \, \Omega \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{30 \, \text{V}}{10 \, \Omega} = 3 \, \text{A}
\]

3. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I_T \cdot R_1 = 3 \, \text{A} \cdot 6 \, \Omega = 18 \, \text{V}
\]

4. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = I_T \cdot R_2 = 3 \, \text{A} \cdot 4 \, \Omega = 12 \, \text{V}
\]

Final Answers for Problem 12:
\[
\boxed{R_T = 10 \, \Omega, \, V_1 = 18 \, \text{V}, \, V_2 = 12 \, \text{V}, \, I_T = 3 \, \text{A}}
\]

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Problem 13

Circuit Description:
- Voltage source \( V_T = 120 \, \text{V} \)
- Resistors \( R_1 = 20 \, \Omega \), \( R_2 \) (unknown), \( R_3 \) (unknown)
- Voltage across \( R_2 \) is given as \( V_2 = 20 \, \text{V} \)
- Voltage across \( R_3 \) is given as \( V_3 = 60 \, \text{V} \)

#### Solution:
1. Current through \( R_2 \) (\( I_2 \)):
Using Ohm's Law for \( R_2 \):
\[
I_2 = \frac{V_2}{R_2}
\]
(We need to find \( R_2 \) later.)

2. Current through \( R_3 \):
Using Ohm's Law for \( R_3 \):
\[
I_3 = \frac{V_3}{R_3}
\]
(We need to find \( R_3 \) later.)

3. Total Current (\( I_T \)):
Since the current splits at the junction:
\[
I_T = I_2 + I_3
\]

4. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = V_T - V_2 - V_3 = 120 \, \text{V} - 20 \, \text{V} - 60 \, \text{V} = 40 \, \text{V}
\]

5. Resistance \( R_1 \):
Using Ohm's Law:
\[
R_1 = \frac{V_1}{I_T}
\]
(We need \( I_T \) to find \( R_1 \).)

6. Resistance \( R_2 \):
Using Ohm's Law:
\[
R_2 = \frac{V_2}{I_2}
\]

7. Resistance \( R_3 \):
Using Ohm's Law:
\[
R_3 = \frac{V_3}{I_3}
\]

Since the problem does not provide enough information to determine \( I_2 \) and \( I_3 \) directly, we assume the currents are equal for simplicity (if not specified otherwise).

Final Answers for Problem 13:
\[
\boxed{V_1 = 40 \, \text{V}, \, I_T = \text{(depends on } I_2 \text{ and } I_3\text{)}, \, R_2 = \text{(depends on } I_2\text{)}, \, R_3 = \text{(depends on } I_3\text{)}}
\]

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Problem 14

Circuit Description:
- Voltage source \( V_T = 150 \, \text{V} \)
- Resistors \( R_1 = 25 \, \Omega \), \( R_2 \) (unknown)
- Voltage across \( R_2 \) is given as \( V_2 = 75 \, \text{V} \)

#### Solution:
1. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = V_T - V_2 = 150 \, \text{V} - 75 \, \text{V} = 75 \, \text{V}
\]

2. Current through \( R_1 \) (\( I_1 \)):
Using Ohm's Law:
\[
I_1 = \frac{V_1}{R_1} = \frac{75 \, \text{V}}{25 \, \Omega} = 3 \, \text{A}
\]

3. Resistance \( R_2 \):
Using Ohm's Law:
\[
R_2 = \frac{V_2}{I_1} = \frac{75 \, \text{V}}{3 \, \text{A}} = 25 \, \Omega
\]

Final Answers for Problem 14:
\[
\boxed{V_1 = 75 \, \text{V}, \, R_2 = 25 \, \Omega, \, I_1 = 3 \, \text{A}}
\]

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Summary of Final Answers:

1. Problem 7: \( V_1 = 25 \, \text{V}, \, V_2 = 75 \, \text{V}, \, I_T = 1.25 \, \text{A} \)
2. Problem 8: \( V_1 = 40 \, \text{V}, \, V_2 = 120 \, \text{V}, \, V_3 = 60 \, \text{V}, \, R_T = 110 \, \Omega \)
3. Problem 9: \( R_2 = 10 \, \Omega, \, V_1 = 40 \, \text{V}, \, V_T = 80 \, \text{V} \)
4. Problem 10: \( V_1 = 30 \, \text{V}, \, V_T = 90 \, \text{V}, \, R_2 = 20 \, \Omega \)
5. Problem 11: \( I_T = 3 \, \text{A}, \, V_1 = 24 \, \text{V} \)
6. Problem 12: \( R_T = 10 \, \Omega, \, V_1 = 18 \, \text{V}, \, V_2 = 12 \, \text{V}, \, I_T = 3 \, \text{A} \)
7. Problem 13: \( V_1 = 40 \, \text{V}, \, I_T = \text{(depends on } I_2 \text{ and } I_3\text{)}, \, R_2 = \text{(depends on } I_2\text{)}, \, R_3 = \text{(depends on } I_3\text{)} \)
8. Problem 14: \( V_1 = 75 \, \text{V}, \, R_2 = 25 \, \Omega, \, I_1 = 3 \, \text{A} \)

\boxed{\text{See detailed solutions above for each problem.}}