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Then, the Thevenin source V 0 Vo:c: I3 5 Vin =41, and RTh Vo:c: 150 Is:c: 41 With RL connected to terminals ab, the output

String names[] = { "Ann", "Bob", "Cal", Vector<String> v(names);

Use superposition to nd the current I from each voltage source in the circuit shown in Fig. 4-30.

"Dan", "Eve"

CHAP. 4]

Example 2.4 Finding the Maximum of Three Integers This program prints the largest of the three numbers input: int n1, n2, n3; cout <<"Enter three integers: "; cin >>n1 >>n2 >>n3; int max=n1; if (n2>max) max=n2; if (n3>max) max=n3; cout <<"The maximum is " <<max <<endl;

13.20 Add the following two member functions to the Vet tor class template: T min(); T max(>; // returns the minimum element in the vector // returns the minimum element in the vector

Fig. 4-29

Use of these functions, of course, presume that the order operators <, >, etc., are defined for the type T. For example, they would not work on an instance of Vector<Complex>.

From the 460-V source, I10 I 0 and for the 200-V source I100 I 00 Then, 200 27 1:65 A 3267 460 74 10:42 A 3267

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13.21 Add the following member function to the Vet t or class template: int find(T t);

Fig. 4-31(a)

This function searches for the element t in the Vet tor. If found, it returns the index of the element; otherwise, it returns -1. Use the sequential search algorithm.

class template:

As a rst step, two-resistor parallel combinations are converted to their equivalents. For the 6 and 3 , Req 6 3 = 6 3 2 . For the two 4- resistors, Req 2 . The circuit is redrawn with series resistors added [Fig. 4-31(b)]. Now the two 6- resistors in parallel have the equivalent Req 3 , and this is in series with the 2 . Hence, RT 5 , as shown in Fig. 4-31(c). The resulting total current is

On the first run, n1 is 22, n2 is 44, and n3 is 66. First max is assigned 22. Then, since 44 > 22, max is assigned 44. Finally, since 66 > 44, max is assigned 66, and that value is printed. On the second run, n1 is 77, n2 is 33, and n3 is 55. First max is assigned 77. Then, since 33 is not greater than 77, max is unchanged. Finally, since 55 is also not greater than 77, max is again unchanged, and so the value 77 is printed.

ostream&

Now the branch currents can be obtained by working back through the circuits of Fig. 4-31(b) and 4-31(a) IC IF 1 IT 2:5 A 2 ID IE 1 IC 1:25 A 2 3 5 I A 6 3 T 3 6 10 IB I A 6 3 T 3 IA

friend istream&

Find the value of the adjustable resistance R which results in maximum power transfer across the terminals ab of the circuit shown in Fig. 4-32.

operator<< ostream&, operator>> istream&,

Fig. 4-32 First a Thevenin equivalent is obtained, with V 0 60 V and R 0 11 . power transfer occurs for R R 0 11 , with Pmax V 02 81:82 W 4R 0 By Section 4.10, maximum

const Vector<T>&); Vector<T>&);

4.17 Apply the mesh current method to the network of Fig. 4-33 and write the matrix equations by inspection. Obtain current I1 by expanding the numerator determinant about the column containing the voltage sources to show that each source supplies a current of 2.13 A.

13.23 Add the following member function to the Vet t or class template: void sort();

page_19

CHAP. 4]

This function rearranges the elements of the Vet t or so that they are in increasing order. Use your favorite sorting algorithm. Use of this function, of course, requires that the order operators C, >, etc., are defined for the type T.

Fig. 4-33 4.18 Loop currents are shown in the network of Fig. 4-34. currents. Ans. 3.55 A, 1:98 A, 2:98 A Write the matrix equation and solve for the three

13.24 Add the following member function to the Vet t or class template: int find(T t);

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