Series and parallel resistor circuit examples with formulas and calculated total resistance values.

To solve the problem, we need to calculate the equivalent resistance for each of the given circuits. Let's go through each case step by step.

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Case 1: Series Circuit

The resistors are connected in series. The formula for the equivalent resistance \( R_{\text{eq}} \) in a series circuit is:

\[
R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots
\]

Given:
- \( R_1 = 8 \, \Omega \)
- \( R_2 = 2 \, \Omega \)
- \( R_3 = 4 \, \Omega \)

\[
R_{\text{eq}} = 8 + 2 + 4 = 14 \, \Omega
\]

Answer for Case 1:
\[
\boxed{14 \, \Omega}
\]

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Case 2: Parallel Circuit

The resistors are connected in parallel. The formula for the equivalent resistance \( R_{\text{eq}} \) in a parallel circuit is:

\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots
\]

Given:
- \( R_1 = 4 \, \Omega \)
- \( R_2 = 6 \, \Omega \)

\[
\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6}
\]

Find a common denominator (which is 12):

\[
\frac{1}{R_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}
\]

Thus,

\[
R_{\text{eq}} = \frac{12}{5} = 2.4 \, \Omega
\]

Answer for Case 2:
\[
\boxed{2.4 \, \Omega}
\]

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Case 3: Mixed Circuit (Series and Parallel)

The circuit consists of a combination of series and parallel connections. We will simplify it step by step.

#### Step 1: Simplify the parallel section
The two resistors in parallel are:
- \( R_1 = 2 \, \Omega \)
- \( R_2 = 4 \, \Omega \)

Using the parallel formula:

\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{4}
\]

Find a common denominator (which is 4):

\[
\frac{1}{R_{\text{parallel}}} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}
\]

Thus,

\[
R_{\text{parallel}} = \frac{4}{3} \, \Omega
\]

#### Step 2: Add the series resistor
The simplified parallel resistance \( R_{\text{parallel}} = \frac{4}{3} \, \Omega \) is now in series with another resistor \( R_3 = 8 \, \Omega \).

Using the series formula:

\[
R_{\text{eq}} = R_{\text{parallel}} + R_3 = \frac{4}{3} + 8
\]

Convert 8 to a fraction with a denominator of 3:

\[
R_{\text{eq}} = \frac{4}{3} + \frac{24}{3} = \frac{28}{3} \, \Omega
\]

Answer for Case 3:
\[
\boxed{\frac{28}{3} \, \Omega}
\]

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Case 4: Complex Mixed Circuit

This circuit involves multiple series and parallel combinations. We will simplify it step by step.

#### Step 1: Simplify the inner parallel section
The two resistors in parallel are:
- \( R_1 = 4 \, \Omega \)
- \( R_2 = 6 \, \Omega \)

Using the parallel formula:

\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{6}
\]

Find a common denominator (which is 12):

\[
\frac{1}{R_{\text{parallel}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}
\]

Thus,

\[
R_{\text{parallel}} = \frac{12}{5} \, \Omega
\]

#### Step 2: Add the series resistor
The simplified parallel resistance \( R_{\text{parallel}} = \frac{12}{5} \, \Omega \) is now in series with another resistor \( R_3 = 8 \, \Omega \).

Using the series formula:

\[
R_{\text{series}} = R_{\text{parallel}} + R_3 = \frac{12}{5} + 8
\]

Convert 8 to a fraction with a denominator of 5:

\[
R_{\text{series}} = \frac{12}{5} + \frac{40}{5} = \frac{52}{5} \, \Omega
\]

#### Step 3: Simplify the outer parallel section
The two resistors in parallel are:
- \( R_1 = 2 \, \Omega \)
- \( R_2 = \frac{52}{5} \, \Omega \)

Using the parallel formula:

\[
\frac{1}{R_{\text{eq}}} = \frac{1}{2} + \frac{1}{\frac{52}{5}}
\]

Simplify \( \frac{1}{\frac{52}{5}} \):

\[
\frac{1}{\frac{52}{5}} = \frac{5}{52}
\]

Thus,

\[
\frac{1}{R_{\text{eq}}} = \frac{1}{2} + \frac{5}{52}
\]

Find a common denominator (which is 52):

\[
\frac{1}{R_{\text{eq}}} = \frac{26}{52} + \frac{5}{52} = \frac{31}{52}
\]

Thus,

\[
R_{\text{eq}} = \frac{52}{31} \, \Omega
\]

Answer for Case 4:
\[
\boxed{\frac{52}{31} \, \Omega}
\]

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Final Answers:

1. \(\boxed{14 \, \Omega}\)
2. \(\boxed{2.4 \, \Omega}\)
3. \(\boxed{\frac{28}{3} \, \Omega}\)
4. \(\boxed{\frac{52}{31} \, \Omega}\)