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2.34 A Si diode has a saturation currrent Io 10 nA at T 3008K. (a) Find the forward current iD if the forward drop vD is 0.5 V. (b) This diode is rated for a maximum current of 5 A. What is its junction temperature at rated current if the forward drop is 0.7 V. Ans: a 2:47 A; (b) 405.48K Solve Problem 2.1 for a Si diode. Ans: vD > :0:2372 V

SellerID Name Address Fee(%)

12.1 To insert a new record with key 16 into the tree shown in Figure 12.16, the initial search would lead to the first leaf node. Since that is a five-way search tree, that first leaf node has overflowed, causing it to be split into two leaf nodes and moving its middle key 19 up to its parent node, as shown in Figure 12.17. But now that parent node has overflowed. So it also gets split, moving its middle key up to its parent node, as shown in Figure 12.18. Two other ordering of the seven keys in Example 12.5 on page 235 that will produce the same BST: a. 44, 22, 33, 77, 55, 99, 88 b. 44, 22, 77, 33, 55, 99, 88 An array of objects could be sorted by inserting their objects into a binary search tree and then using an inorder traversal to copy them back into the array. The BST property guarantees that the inorder traversal will visit the elements in order. If an AVL tree is used, then each insertion runs in O(lgn) time, so building the tree with n elements will require O(n lgn) time. The subsequent inorder traversal also has O(n lgn) complexity, so the entire algorithm sorts the array in O(n lgn) time. All except a are binary search trees.

2.35 2.36

private AVLTree rotateRight() { AVLTree x = this, y = left, z = y.left; x.left = z; y.left = x; int xb = x.balance; int yb = y.balance;

Laboratory data for a Si diode described by (2.1) show that iD 2 mA when vD 0:6 V, and iD 10 mA for vD 0:7 V. Find (a) the temperature for which the data were taken, and (b) the reverse saturation current. Ans: a 87:198C; b 2:397 A For what voltage vD will the reverse current of a Ge diode that is described by (2.1) reach 99 percent of its saturation value at a temperature of 3008K Ans: vD 0:1191 V Find the increase in temperature T necessary to increase the reverse saturation current of a diode by a factor of 100. Ans: 66:48C

Delivery_Services PK ServiceID Name Address Phone PrimaryContact StoreExpense PK PK StoreExpenseType Date Payee Amount CheckNo ExpenseType Rent, heat, etc. StoreExpenseType PK ExpenseType PK PK FK1 FK2

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The diode of Problem 2.34 is operating in a circuit where it has dynamic resistance rd 100 . What must be the quiescent conditions Ans: VDQ 0:263 V; IDQ 0:259 mA The diode of Problem 2.34 has a forward current iD 2 0:004 sin !t mA. Ans: vD 339:5 0:0207 sin !t mV vD VDQ vd , across the diode. Find the total voltage,

if (yb > 1) { ++x.balance; y.balance = ( xb<0 yb+1 : xb+yb+2 ); } else if (yb > xb) { x.balance += yb-1; ++y.balance; } else { y.balance = xb+2; } return y; }

8.3 An inexperienced database designer suggested the following schema for one of the tables in the antique store database: Sales Item_ID | Description | Price | Date | Customer_Name | Addr | City | St | ZIP | Phone

Find the power dissipated in the load resistor RL 100 of the circuit of Fig. 2-22(a) if the diode is ideal and vS 10 sin !t V. Ans: 206:6 mW The logic AND gate of Fig. 2-46(a) has trains of input pulses arriving at the gate inputs, as indicated by Fig. 2-47(b). Signal v2 is erratic, dropping below nominal logic level on occasion. Determine vo . Ans: 10 V for 1 t 2 ms, 5 V for 4 t 5 ms, zero otherwise.

Theorem. Every subtree of a binary search tree is a binary search tree. Proof: Let T be a binary search tree, and let S be a subtree of T. Let x be any element in S, and let L and R be the left and right subtrees of x in S. Then, since S is a subtree of T, x is also an element of T, and L and R are the left and right subtrees of x in T. Therefore, y x z for every y L and every z R because T has the BST property. Thus, S also has the BST property.

Fig. 2-46