Given:

• Terminal voltage \(V_1 = 30\,\text{V}\) when delivering \(I_1 = 400\,\text{mA} = 0.4\,\text{A}\)
• Terminal voltage \(V_2 = 28\,\text{V}\) when delivering \(I_2 = 600\,\text{mA} = 0.6\,\text{A}\)

Step 1: Find Internal Resistance

For a practical voltage source (Thevenin form):

\(V = V_{th} - IR_{th}\)

Using the given values:

\(30 = V_{th} - 0.4R_{th}\)
\(28 = V_{th} - 0.6R_{th}\)

Subtracting the two equations:

\(30 - 28 = (V_{th} - 0.4R_{th}) - (V_{th} - 0.6R_{th})\)
\(2 = 0.2R_{th}\)
\(\therefore R_{th} = 10\,\Omega\)

Step 2: Find Thevenin Voltage

Using:

\(30 = V_{th} - 0.4(10)\)
\(30 = V_{th} - 4\)
\(\therefore V_{th} = 34\,\text{V}\)

Step 3: Convert Thevenin Equivalent to Norton Equivalent

For Thevenin-to-Norton conversion:

\(R_N = R_{th}\)
\(I_N = \dfrac{V_{th}}{R_{th}}\)

Therefore:

\(R_N = 10\,\Omega\)
\(I_N = \dfrac{34}{10} = 3.4\,\text{A}\)

Final Norton Equivalent:

• Norton Current: \(I_N = 3.4\,\text{A}\)
• Norton Resistance: \(R_N = 10\,\Omega\)

Equivalent Circuit:

The Norton equivalent circuit consists of a 3.4 A current source connected in parallel with a 10 Ω resistor.

Final Answer:

\[ \boxed{I_N = 3.4\,\text{A}, \qquad R_N = 10\,\Omega} \]

3-input XOR gate using 2:1 muxes: A 3-input XOR gate returns "1" when the odd number of inputs are "1". The truth table for 3-input XOR gate is shown in figure below.

Truth table for 3-input XOR gate Let us choose A to be the select of mux closest to output. When A is "0", output is a function of B and C. So, we need another mux, let us choose B to be the select of this mux. Among the rows with A = 0, when B is "0", output is equal to C and when B is "1", output is equal to C'.

Similarly among the rows with A = "1", when B is "0", output is equal to C' and when B is "1", output is equal to C. The implementation of 3-input XOR gate using 2:1 muxes is shown in figure below.

Given:

• Logical Address Size = 16 bits
• Page Size = 1 KB = 2¹⁰ bytes

(i) How many pages are in the logical address space?

First, calculate the total logical address space:

Total Address Space = 2¹⁶ bytes

Since each page contains 2¹⁰ bytes:

Number of Pages = 2¹⁶ / 2¹⁰

= 2^{16 − 10}

= 2⁶ = 64 pages

Therefore, the logical address space contains 64 pages.

(ii) How many bits are used for the page number and offset?

The logical address is divided into two parts:

• Page Number: Identifies which page the address belongs to.
• Offset: Identifies the exact location within that page.

Since there are 64 pages:

Page Number Bits = log₂(64) = 6 bits

Since the page size is 1 KB = 2¹⁰ bytes:

Offset Bits = log₂(2¹⁰) = 10 bits

Therefore:

• Page Number = 6 bits
• Page Offset = 10 bits
• Total = 6 + 10 = 16 bits

Final Answer:

• Number of Pages = 64
• Page Number Bits = 6
• Offset Bits = 10

Command:

head -10 wasacustomers.txt | wc -mw

Explanation:

• head -10 wasacustomers.txt → Extracts the first 10 lines of the file.
• wc -m → Counts the total number of characters.
• wc -w → Counts the total number of words.
• | → Pipes the output of the head command to wc.

Therefore:

head -10 wasacustomers.txt | wc -mw

This command counts the characters and words from the first 10 lines of wasacustomers.txt.

Step 1: Evaluate i++

Initially, i = -1. Since -1 is a non-zero value in C, it is considered true.

Therefore, i++ evaluates to true, and then i becomes 0.

Step 2: Evaluate j++

Since the left side of && is true, j++ is evaluated. Initially, j = -1, which is also true. After post-increment, j becomes 0.

Step 3: Evaluate k++

Since both previous operands are true, k++ is evaluated. Initially, k = 0, which is false. After post-increment, k becomes 1.

Therefore:

i++ && j++ && k++ → false

Step 4: Evaluate l++

Because the left side of || is false, the right side l++ must be evaluated. Initially, l = 2, so l++ returns 2 (true), then l becomes 3.

Thus:

false || true → true

Therefore, m = 1.

Final Values:

• i = 0
• j = 0
• k = 1
• l = 3
• m = 1

Output:

0 0 1 3 1

Key Concept: C uses short-circuit evaluation for && and ||. Also, a post-increment operator (++) uses the original value first and increments the variable afterward.

Assume the Employee table contains the following columns:

• employee_id — Employee ID
• employee_name — Employee Name
• join_date — Joining Date
• salary — Employee Salary

(a) Find the employees who joined on the same date.

First, identify the joining dates that occur more than once, then retrieve the employees having those dates:

SELECT *
FROM Employee
WHERE join_date IN (
    SELECT join_date
    FROM Employee
    GROUP BY join_date
    HAVING COUNT(*) > 1
);

Explanation:

• GROUP BY join_date → Groups employees according to their joining date.
• HAVING COUNT(*) > 1 → Selects dates on which more than one employee joined.
• The outer query displays all employees whose joining date matches those dates.

(b) Find employees whose salary is greater than 8,000 and less than 25,000.

SELECT *
FROM Employee
WHERE salary > 8000
  AND salary < 25000;

Explanation:

• salary > 8000 → The salary must be greater than 8,000.
• salary < 25000 → The salary must be less than 25,000.
• AND → Both conditions must be true.

Alternative for (b):

SELECT *
FROM Employee
WHERE salary BETWEEN 8001 AND 24999;

Important: BETWEEN is inclusive. Therefore, for the exact condition greater than 8,000 and less than 25,000, using > 8000 and < 25000 is clearer and safer.

Given Network:

• Network Address: 245.248.128.0/20
• Subnet Mask: 255.255.240.0
• Total Addresses: 2^32-20 = 2¹² = 4096

The address space needs to be divided among three departments:

• Manager: Half of the address space = 50% = 2048 addresses
• HR: One-quarter of the address space = 25% = 1024 addresses
• Admin: Remaining one-quarter = 25% = 1024 addresses

Step 1: Manager — Half of the Address Space

Manager needs 2048 addresses.

2048 = 2¹¹

Therefore, the new prefix is:

32 − 11 = /21

• Network: 245.248.128.0/21
• Subnet Mask: 255.255.248.0
• Address Range: 245.248.128.0 – 245.248.135.255
• Usable Host Range: 245.248.128.1 – 245.248.135.254
• Broadcast: 245.248.135.255

Step 2: HR — One-Quarter of the Address Space

HR needs 1024 addresses.

1024 = 2¹⁰

Therefore, the new prefix is:

32 − 10 = /22

• Network: 245.248.136.0/22
• Subnet Mask: 255.255.252.0
• Address Range: 245.248.136.0 – 245.248.139.255
• Usable Host Range: 245.248.136.1 – 245.248.139.254
• Broadcast: 245.248.139.255

Step 3: Admin — Remaining One-Quarter

Admin also needs 1024 addresses.

1024 = 2¹⁰

Therefore, the new prefix is /22.

• Network: 245.248.140.0/22
• Subnet Mask: 255.255.252.0
• Address Range: 245.248.140.0 – 245.248.143.255
• Usable Host Range: 245.248.140.1 – 245.248.143.254
• Broadcast: 245.248.143.255

Final Subnet Allocation:

Final Answer:

Manager gets 245.248.128.0/21, HR gets 245.248.136.0/22, and Admin gets 245.248.140.0/22.