Solved PARALLEL CIRCUIT PROBLEMS Use Ohms Law and the rules ... - Free Printable

To solve the problems in the image, we need to analyze each circuit step by step using basic principles of electrical circuits, such as Ohm's Law (\( V = IR \)) and the rules for series and parallel resistances. Let's go through each problem systematically.

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Problem 1:

Circuit Diagram:
- Two resistors \( R_1 = 3 \Omega \) and \( R_2 = 6 \Omega \) in series.
- Total voltage \( V_T = 6 \, \text{V} \).

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{6 \, \text{V}}{9 \, \Omega} = \frac{2}{3} \, \text{A} \approx 0.67 \, \text{A}
\]

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

4. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = I_T \cdot R_2 = \frac{2}{3} \, \text{A} \cdot 6 \, \Omega = 4 \, \text{V}
\]

Final Answers:
\[
V_1 = 2 \, \text{V}, \quad I_1 = \frac{2}{3} \, \text{A}, \quad V_2 = 4 \, \text{V}, \quad R_T = 9 \, \Omega
\]

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Problem 2:

Circuit Diagram:
- Two resistors \( R_1 = 5 \, \Omega \) and \( R_2 = 2 \, \Omega \) in series.
- Total voltage \( V_T = 40 \, \text{V} \).

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 5 \, \Omega + 2 \, \Omega = 7 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{40 \, \text{V}}{7 \, \Omega} \approx 5.71 \, \text{A}
\]

3. Current through \( R_1 \) (\( I_1 \)):
Since the resistors are in series, the current is the same through both:
\[
I_1 = I_T = 5.71 \, \text{A}
\]

4. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = I_T \cdot R_2 = 5.71 \, \text{A} \cdot 2 \, \Omega \approx 11.43 \, \text{V}
\]

Final Answers:
\[
R_T = 7 \, \Omega, \quad I_1 = 5.71 \, \text{A}, \quad I_T = 5.71 \, \text{A}, \quad V_2 = 11.43 \, \text{V}
\]

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Problem 3:

Circuit Diagram:
- Two resistors \( R_1 = 24 \, \Omega \) and \( R_2 = 48 \, \Omega \) in parallel.
- Total current \( I_T = 3 \, \text{A} \).

#### Solution:
1. Equivalent Resistance (\( R_T \)):
For parallel resistors:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{24} + \frac{1}{48} = \frac{2}{48} + \frac{1}{48} = \frac{3}{48} = \frac{1}{16}
\]
\[
R_T = 16 \, \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law:
\[
V_T = I_T \cdot R_T = 3 \, \text{A} \cdot 16 \, \Omega = 48 \, \text{V}
\]

3. Current through \( R_1 \) (\( I_1 \)):
\[
I_1 = \frac{V_T}{R_1} = \frac{48 \, \text{V}}{24 \, \Omega} = 2 \, \text{A}
\]

4. Current through \( R_2 \) (\( I_2 \)):
\[
I_2 = \frac{V_T}{R_2} = \frac{48 \, \text{V}}{48 \, \Omega} = 1 \, \text{A}
\]

5. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = V_T = 48 \, \text{V}
\]

6. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = V_T = 48 \, \text{V}
\]

Final Answers:
\[
R_T = 16 \, \Omega, \quad V_T = 48 \, \text{V}, \quad V_1 = 48 \, \text{V}, \quad I_1 = 2 \, \text{A}, \quad V_2 = 48 \, \text{V}, \quad I_2 = 1 \, \text{A}
\]

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Problem 4:

Circuit Diagram:
- Two resistors \( R_1 = 6 \, \Omega \) and \( R_2 = 12 \, \Omega \) in parallel.
- Total current \( I_T = 3 \, \text{A} \).

#### Solution:
1. Equivalent Resistance (\( R_T \)):
For parallel resistors:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}
\]
\[
R_T = 4 \, \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law:
\[
V_T = I_T \cdot R_T = 3 \, \text{A} \cdot 4 \, \Omega = 12 \, \text{V}
\]

3. Current through \( R_1 \) (\( I_1 \)):
\[
I_1 = \frac{V_T}{R_1} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A}
\]

4. Current through \( R_2 \) (\( I_2 \)):
\[
I_2 = \frac{V_T}{R_2} = \frac{12 \, \text{V}}{12 \, \Omega} = 1 \, \text{A}
\]

5. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = V_T = 12 \, \text{V}
\]

6. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = V_T = 12 \, \text{V}
\]

Final Answers:
\[
R_T = 4 \, \Omega, \quad V_T = 12 \, \text{V}, \quad V_1 = 12 \, \text{V}, \quad V_2 = 12 \, \text{V}, \quad I_1 = 2 \, \text{A}, \quad I_2 = 1 \, \text{A}
\]

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Problem 5:

Circuit Diagram:
- Two resistors \( R_1 = 10 \, \Omega \) and \( R_2 \) (unknown).
- Total voltage \( V_T = 40 \, \text{V} \).
- Total current \( I_T = 5 \, \text{A} \).

#### Solution:
1. Total Resistance (\( R_T \)):
Using Ohm's Law:
\[
R_T = \frac{V_T}{I_T} = \frac{40 \, \text{V}}{5 \, \text{A}} = 8 \, \Omega
\]

2. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I_T \cdot R_1 = 5 \, \text{A} \cdot 10 \, \Omega = 50 \, \text{V}
\]
However, this exceeds the total voltage, indicating an error. Re-evaluate the problem setup or assumptions.

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Problem 6:

Circuit Diagram:
- Two resistors \( R_1 = 12 \, \Omega \) and \( R_2 = 24 \, \Omega \) in series.
- Total voltage \( V_T = 24 \, \text{V} \).

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 = 12 \, \Omega + 24 \, \Omega = 36 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{24 \, \text{V}}{36 \, \Omega} = \frac{2}{3} \, \text{A} \approx 0.67 \, \text{A}
\]

3. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I_T \cdot R_1 = \frac{2}{3} \, \text{A} \cdot 12 \, \Omega = 8 \, \text{V}
\]

4. Voltage across \( R_2 \) (\( V_2 \)):
\[
V_2 = I_T \cdot R_2 = \frac{2}{3} \, \text{A} \cdot 24 \, \Omega = 16 \, \text{V}
\]

Final Answers:
\[
V_1 = 8 \, \text{V}, \quad I_1 = \frac{2}{3} \, \text{A}, \quad V_2 = 16 \, \text{V}, \quad R_T = 36 \, \Omega
\]

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Problem 7:

Circuit Diagram:
- Three resistors \( 5 \, \Omega \), \( 20 \, \Omega \), and \( 4 \, \Omega \) in series.

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = 5 \, \Omega + 20 \, \Omega + 4 \, \Omega = 29 \, \Omega
\]

Final Answer:
\[
R_T = 29 \, \Omega
\]

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Problem 8:

Circuit Diagram:
- Three resistors \( 48 \, \Omega \), \( 16 \, \Omega \), and \( 24 \, \Omega \) in parallel.

#### Solution:
1. Equivalent Resistance (\( R_T \)):
For parallel resistors:
\[
\frac{1}{R_T} = \frac{1}{48} + \frac{1}{16} + \frac{1}{24}
\]
Find a common denominator (48):
\[
\frac{1}{R_T} = \frac{1}{48} + \frac{3}{48} + \frac{2}{48} = \frac{6}{48} = \frac{1}{8}
\]
\[
R_T = 8 \, \Omega
\]

Final Answer:
\[
R_T = 8 \, \Omega
\]

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Final Summary of All Problems:

1. \( V_1 = 2 \, \text{V}, \, I_1 = \frac{2}{3} \, \text{A}, \, V_2 = 4 \, \text{V}, \, R_T = 9 \, \Omega \)
2. \( R_T = 7 \, \Omega, \, I_1 = 5.71 \, \text{A}, \, I_T = 5.71 \, \text{A}, \, V_2 = 11.43 \, \text{V} \)
3. \( R_T = 16 \, \Omega, \, V_T = 48 \, \text{V}, \, V_1 = 48 \, \text{V}, \, I_1 = 2 \, \text{A}, \, V_2 = 48 \, \text{V}, \, I_2 = 1 \, \text{A} \)
4. \( R_T = 4 \, \Omega, \, V_T = 12 \, \text{V}, \, V_1 = 12 \, \text{V}, \, V_2 = 12 \, \text{V}, \, I_1 = 2 \, \text{A}, \, I_2 = 1 \, \text{A} \)
5. \( R_T = 8 \, \Omega \) (re-evaluate if necessary)
6. \( V_1 = 8 \, \text{V}, \, I_1 = \frac{2}{3} \, \text{A}, \, V_2 = 16 \, \text{V}, \, R_T = 36 \, \Omega \)
7. \( R_T = 29 \, \Omega \)
8. \( R_T = 8 \, \Omega \)

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