KVL at Node \(i_1 = \Sigma V = 0\)

The equation is:

\(-V_{x} + 10 i_{1} + 4 V_{x} + 15 i_{1} = 0\)

\(25 i_{1} + 3 (-5i_{1}) = 0\)

\(i_{1} = 0 \, \text{amp}\)

The voltage across terminals \(V_{ab}\) is:

\(V_{ab} = 15i_{1} = 0V\)

Finally, the Thevenin voltage:

\(V_{TH} = 0V\)

KVL at \(i_{1}: \Sigma V = 0\)

The equation is:

\(-V_{x} + 10 i_{1} + 4 V_{x} + 15 (i_1 + i_{0}) = 0\)

Simplifying:

\(3 V_{x} + 25 i_{1} + 15 i_{0}\) = 0

=>\(3 (-5 i_{1}) + 25 i_{1} + 15 i_{0}\) = 0

=>\(15 i_{0} + 10 i_{1} = 0 \quad \dots \text{(1)}\)

KVL at \(i_{0}: \Sigma V = 0\)

The equation is:

\(-V_{x} + 10 i_{1} + 4 V_{x} + 15 (i_1 + i_{0}) = 0\)

Simplifying:

\(-1 + 15 (i_{1} + i_{0}) \) = 0

=>\(15 i_{0} + 15 i_{1} = 1 \quad \dots \text{(2)}\)

solving (1,2) i_o = – 2/15 amp

So R_th= 1/i_o= -15/2 = -7.5 ohm

Voltage at Point a:
V_a = -(6/2) V₁
V_a = -3 V₁

Voltage at Point b:
V_b = -(8/4)V₂
V_b = -2V₂

Summing amplifier:
V₀ = -(10/5)V_a-(10/15)V_b
V₀=-2V_a – (2/3)V_b
V₀=-2(-3V₁) -(2/3)(-2V₂)
V₀ = 6 V₁ + (4/3) V₂
= 8.667

#include <stdio.h>

// Function to sort the array using simple bubble sort
void sort(int arr[], int n) {
    int i, j, temp;
    for(i = 0; i < n-1; i++) {
        for(j = 0; j < n-i-1; j++) { if(arr[j] > arr[j+1]) {
                temp = arr[j];
                arr[j] = arr[j+1];
                arr[j+1] = temp;
            }
        }
    }
}

int main() {
    int n, k;

    printf("Enter number of elements: ");
    scanf("%d", &n);

    int arr[n];

    printf("Enter %d elements:\n", n);
    for(int i = 0; i < n; i++) { scanf("%d", &arr[i]); } printf("Enter value of k: "); scanf("%d", &k); if(k > n) {
        printf("k must be less than or equal to n\n");
        return 0;
    }

    // Sort the array
    sort(arr, n);

    // kth smallest element
    printf("The %dth smallest element is: %d\n", k, arr[k-1]);

    return 0;
}

Sample Input:
Enter number of elements: 5
Enter 5 elements:
12 5 7 3 9
Enter value of k: 2

Sample Output:
The 2th smallest element is: 5

-- 1. Find the firstname of customers who order all products
--    where Category = 'B' and OrderDate = '15 March 2024'

SELECT DISTINCT c.FirstName
FROM Customer c
JOIN Orders o ON c.CustomerID = o.CustomerID
JOIN Product p ON o.ProductID = p.ProductID
WHERE p.Category = 'B'
  AND o.OrderDate = '2024-03-15'
  AND NOT EXISTS (
      SELECT *
      FROM Product p2
      WHERE p2.Category = 'B'
        AND p2.ProductID NOT IN (
            SELECT o2.ProductID
            FROM Orders o2
            WHERE o2.CustomerID = c.CustomerID
              AND o2.OrderDate = '2024-03-15'
        )
  );

-- 2. Find the TotalOrder as Monetary Amount
--    which is ordered between 15 March 2024 and 29 March 2024

SELECT SUM(op.Quantity * p.Price) AS TotalOrder
FROM Orders o
JOIN OrderProduct op ON o.OrderID = op.OrderID
JOIN Product p ON op.ProductID = p.ProductID
WHERE o.OrderDate BETWEEN '2024-03-15' AND '2024-03-29';