16. In the differentiating circuit given in the figure, function of R₁ is to
    
    

1. 1. enable circuit to approach ideal differentiator

   
   
   

   2. maintain high input impedance

   
   
   

   3. eliminate high frequency noise spikes

   
   
   

   4. prevent oscillations at high frequency

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

17. Expression for the output voltage V_o in terms of the input voltage V₁ and V₂ in the circuit shown in the figure, assuming operational amplifier to be ideal is V_o = A₁ V₁ + A₂ V₂ Values of A₁ and A₂ would be respectively
    
    

1. 1. 9 and – 10

   
   
   

   2. 9.9 and – 10

   
   
   

   3. – 9 and 10

   
   
   

   4. – 9.9 and 10

   
   
   

---

1. View Hint View Answer Discuss in Forum

   VB = V1 -	V1	× 11 = 0.9 V1
   	110

   
   V_A = V_B = 0.9 V₁
   

   Again	V0 - VA	=	VA - V2
   	100 K		10 K

   
   

   ⇒	V0 - 0.9 V1	=	0.9 V1 - V2
   	100		10

   
   ⇒ V₀ = 9.9 V₁ – 10 V₂
   ∴ A₁ = 9.9 and A₂ = – 10

   Correct Option: B

   VB = V1 -	V1	× 11 = 0.9 V1
   	110

   
   V_A = V_B = 0.9 V₁
   

   Again	V0 - VA	=	VA - V2
   	100 K		10 K

   
   

   ⇒	V0 - 0.9 V1	=	0.9 V1 - V2
   	100		10

   
   ⇒ V₀ = 9.9 V₁ – 10 V₂
   ∴ A₁ = 9.9 and A₂ = – 10

---

---

18. An op-amp is open-loop gain of 10⁵ and open-loop upper cut-off frequency of 10 Hz. If this OP-AMP is connected as an amplifier with a closed-loop gain at 100, then new uper cut-off frequency will be
    

1. 1. 10 Hz

   
   
   

   2. 100 Hz

   
   
   

   3. 10 kHz

   
   
   

   4. 100 kHz

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Upper cut-off frequency at closed loop gain
   = f_o × (1 + βA) = 10 × 100 = 10 kHz
   

   Correct Option: C

   Upper cut-off frequency at closed loop gain
   = f_o × (1 + βA) = 10 × 100 = 10 kHz
   

---

19. The period of the output waveform for the circuit of figure shown, when triggered by a negative pulse is
    
    

1. 1. 0.75 ms

   
   
   

   2. 0.825 ms

   
   
   

   3. 0.525 ms

   
   
   

   4. 1.5 ms

   
   
   

---

1. View Hint View Answer Discuss in Forum

   T = (1.1) R_AC
   = 1.1(7.5 × 10³) (0.1 × 10^{– 6}) = 0.825 ms.

   Correct Option: B

   T = (1.1) R_AC
   = 1.1(7.5 × 10³) (0.1 × 10^{– 6}) = 0.825 ms.

---

---

20. For the oscillator circuit shown in the given figure, expression for the time period of oscillation can be given by (where t = RC)
    
    

1. 1. τ ln 3

   
   
   

   2. 2 τ ln 3

   
   
   

   3. τ ln 2

   
   
   

   4. 2 τ ln 2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   T = 2RC ln	1 + β
   	1 - β

   
   

   Here , β =	1	=	R
   	2		R + R

   
   ∴ T = 2 RC ln 3 = 2 τ ln 3

   Correct Option: B

   T = 2RC ln	1 + β
   	1 - β

   
   

   Here , β =	1	=	R
   	2		R + R

   
   ∴ T = 2 RC ln 3 = 2 τ ln 3

---