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Question 1: 
a. For the following network, calculate the system reliability using the component reliability. 
Assume each component has a reliability of R. b. Determine the system reliability if R = 0.9

Let's first break down the circuit network and calculate the system reliability based on the given configuration.

### Step 1: Analyze the Structure of the Network

This network consists of multiple resistors arranged in a combination of series and parallel components. To find the system reliability, we need to analyze the configuration.

- **Outer Loop:** There are three resistors (R) in series along the bottom of the figure.
- **Middle Parallel Group:** In the middle, there are two resistors (R) in parallel. This parallel combination connects to another resistor (R) that lies in series with them.
- **Top Parallel Group:** There are three resistors (R) in parallel at the top of the network.

### Step 2: Calculate Equivalent Reliability for Parallel and Series Combinations

1. **Bottom Path (Series):**
   The three resistors along the bottom are in series. The reliability of resistors in series is the product of their individual reliabilities:

   \[
   R_{\text{series}} = R \cdot R \cdot R = R^3
   \]

2. **Middle Parallel Group:**
   The two resistors in parallel have an equivalent reliability of:

   \[
   R_{\text{parallel}} = 1 - (1 - R)(1 - R) = 1 - (1 - R)^2
   \]

   After calculating the parallel combination, it is in series with one more resistor. Thus, the total reliability for this middle part is:

   \[
   R_{\text{middle}} = R_{\text{parallel}} \cdot R = \left[ 1 - (1 - R)^2 \right] \cdot R
   \]

3. **Top Path (Parallel):**
   There are three resistors in parallel at the top, with the reliability given by:

   \[
   R_{\text{top}} = 1 - (1 - R)^3
   \]

### Step 3: Total System Reliability

Now, the three sections (bottom series, middle combination, and top parallel) are in parallel with each other. Therefore, the total system reliability \( R_{\text{system}} \) is:

\[
R_{\text{system}} = 1 - \left( (1 - R_{\text{series}}) (1 - R_{\text{middle}}) (1 - R_{\text{top}}) \right)
\]

We will now plug in the expressions for \( R_{\text{series}} \), \( R_{\text{middle}} \), and \( R_{\text{top}} \) to compute the total system reliability.

### Step 4: Substitute \( R = 0.9 \) into the Expression

After calculating the symbolic expression for system reliability, we will substitute \( R = 0.9 \) to find the numeric value of system reliability.

Let me compute this for you.The system reliability for the given network is expressed symbolically as:

\[
R_{\text{system}} = 1 - (1 - R^3)(1 - [R(1 - (1 - R)^2)])(1 - (1 - R)^3)
\]

For \( R = 0.9 \), the system reliability is approximately:

\[
R_{\text{system}} \approx 0.99997
\]

This means the network's reliability is about 99.997% when each component has a reliability of 0.9.

Would you like further clarification or details?

### Related Questions:
1. How does the system reliability change if the component reliability decreases to 0.8?
2. What is the impact on system reliability if one resistor in the middle parallel section fails?
3. Can you explain the difference in behavior between series and parallel reliability configurations?
4. How would the system reliability be affected if we added another resistor in series with the middle section?
5. What would happen to the reliability if all resistors were placed in a purely parallel configuration?

### Tip:
In parallel circuits, adding more components increases the overall reliability, whereas in series circuits, each additional component decreases it.

For the following network, calculate the system reliability using the component reliability. Assume each component has a reliability of R. Determine the system reliability if R = 0.9.

Learn how to calculate the system reliability of a complex network of resistors in series and parallel. This problem involves calculating the total reliability using the given reliability of R for each component. Includes step-by-step calculations for reliability, particularly when R = 0.9. Ideal for engineering students.

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