8. Find equivalent resistance across a-b, b-c, c-d and a-c terminals. b. 10 Ω Μ 450 Ω Μ 300 Ω 450 Ω 3300 Ω Μ 60 Ω d

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**Problem 8:**

Find the equivalent resistance across the following terminals:
- a-b
- b-c
- c-d
- a-c

**Diagram Explanation:**

This diagram represents a network of resistors connected in a combination of series and parallel configurations. Here's a detailed breakdown:

- **Between terminals a and c**: There is a series connection of two resistors: 10 Ω and 450 Ω.
- **Between terminals b and d**: There is a series connection of two resistors: 450 Ω and 60 Ω.
- **There are two parallel branches** connecting the central nodes of the above series connections:
  - Both parallel branches have a resistor with resistance of 300 Ω each.

To find the equivalent resistance across different pairs of terminals, you would need to use the rules for series and parallel resistances:
- Resistances in series: \( R_{\text{eq}} = R_1 + R_2 + \ldots + R_n \)
- Resistances in parallel: \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \)

The combination of series and parallel configurations in the network requires step-by-step calculations for each pair of terminals specified.
Transcribed Image Text:**Problem 8:** Find the equivalent resistance across the following terminals: - a-b - b-c - c-d - a-c **Diagram Explanation:** This diagram represents a network of resistors connected in a combination of series and parallel configurations. Here's a detailed breakdown: - **Between terminals a and c**: There is a series connection of two resistors: 10 Ω and 450 Ω. - **Between terminals b and d**: There is a series connection of two resistors: 450 Ω and 60 Ω. - **There are two parallel branches** connecting the central nodes of the above series connections: - Both parallel branches have a resistor with resistance of 300 Ω each. To find the equivalent resistance across different pairs of terminals, you would need to use the rules for series and parallel resistances: - Resistances in series: \( R_{\text{eq}} = R_1 + R_2 + \ldots + R_n \) - Resistances in parallel: \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \) The combination of series and parallel configurations in the network requires step-by-step calculations for each pair of terminals specified.
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