146. Find the output of given circuit—
     
     
     

1. 1. AB + C

   
   
   

   2. AB

   
   
   

   3. A + BC

   
   
   

   4. AC + B

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   
   

   Correct Option: D

   NA

   
   

---

147. Determine the values of A, B, C that makes the sum term A + B + C equal to 0—

1. 1. A = 1, B = 0, C = 0

   
   
   

   2. A = 1, B = 1, C = 0

   
   
   

   3. A = 0, B = 0, C = 1

   
   
   

   4. All of them

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A + B + C = D + 0 + 1
   = 1 + 1 + 1
   = 1
   A = 0, B = 0, C = 1.

   Correct Option: C

   A + B + C = D + 0 + 1
   = 1 + 1 + 1
   = 1
   A = 0, B = 0, C = 1.

---

---

148. Determine the values of A, B, C, D that make the product term – AB– CD equal to 1—

1. 1. A = 0, B = 1, C = 0, D = 1

   
   
   

   2. A = 0, B = 1, C = 1, D = 1

   
   
   

   3. A = 1, B = 1, C = 1, D = 0

   
   
   

   4. A = 1, B = 0, C = 0, D = 0

   
   
   

---

1. View Hint View Answer Discuss in Forum

   AB CD = 0·1·0·1
   = 1·1·1·1
   = 1
   A = 0, B = 1, C = 0, D = 1.

   Correct Option: A

   AB CD = 0·1·0·1
   = 1·1·1·1
   = 1
   A = 0, B = 1, C = 0, D = 1.

---

149. A positive going pulse is applied to an inverter. The time interval from the leading edge of the input to the leading edge of the output is 7 ns. This parameter is—

1. 1. Speed power product

   
   
   

   2. Propagation delay t_PHL

   
   
   

   3. Propagation delay t_PLH

   
   
   

   4. Pulse width

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

---

150. A pulse is applied to each input of a 2 input NAND gate. One pulse goes HIGH at t = 0 and goes back low at t = 1 ms. The other pulse goes HIGH at t = 0·8 ms and goes back low at t = 3 ms. The output pulse can be described as follows—

1. 1. It goes LOW at t = 0 and back HIGH at t = 3 ms

   
   
   

   2. It goes LOW at t = 0·8 ms and back HIGH at t = 3 ms

   
   
   

   3. It goes LOW at t = 0·8 ms and back HIGH at t = 1 ms

   
   
   

   4. It goes LOW at t = 0·8 ms and back LOW at t = 1 ms

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   
   

   Correct Option: C

   NA

   
   

---