411. Among the three configurations the highest output impedance is obtained by

1. 1. CE

   
   
   

   2. CC

   
   
   

   3. CB

   
   
   

   4. CE and CB

   
   
   

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   NA

   Correct Option: C

   NA

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412. Regarding CC amplifier consider the statements 1 to 4—
     1. It perfoms a resistance transfomation from low to high resistance.
     2. Its current gain is close to unity
     3. Its voltage gain is close to unity
     4. The frequency range is higher than that of a CE syage.
     Of these the only true statements are:

1. 1. 1, 2, 4

   
   
   

   2. 1, 3

   
   
   

   3. 2, 4

   
   
   

   4. 3, 4

   
   
   

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   (i) Common collector amplifier perform a resistance transformation from low resistance to high resistance.
   (ii) Its current gain is very high.
   (iii) Its voltage gain close to unity.

   Correct Option: B

   (i) Common collector amplifier perform a resistance transformation from low resistance to high resistance.
   (ii) Its current gain is very high.
   (iii) Its voltage gain close to unity.

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413. In an emitter follwer with R_L =10 K, given h_fe = 99, h_oe = (1/40) × 10^–3, h_{i e} = 1 K the values of current gain and input resistance are given by—

1. 1. 25 and 250 K

   
   
   

   2. 80 and 800 K

   
   
   

   3. 99 and 900 K

   
   
   

   4. 100 and 100 K

   
   
   

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1. View Hint View Answer Discuss in Forum

   In the common emitter amplifier, the current gain is given by the relation,
   

   AI =	–hfe
   	1 + hoe · RL

   
   

   or | AI | =		99	=	99
   		1 + (1 / 40) × 10–3 × 10 × 103		1 + (10 / 40)

   
   

   =	99 x 40	= 79.2 ≅ 80
   	50

   
   R_i = h_ie + A_I h_reR_L
   = 1K + 80 × 1 × 10K
   ≅ 800 K

   Correct Option: B

   In the common emitter amplifier, the current gain is given by the relation,
   

   AI =	–hfe
   	1 + hoe · RL

   
   

   or | AI | =		99	=	99
   		1 + (1 / 40) × 10–3 × 10 × 103		1 + (10 / 40)

   
   

   =	99 x 40	= 79.2 ≅ 80
   	50

   
   R_i = h_ie + A_I h_reR_L
   = 1K + 80 × 1 × 10K
   ≅ 800 K

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414. In the common emitter amplifier with RL = 4000Ω, given hfe = 100, hoe =	1	, hie = 1000, the current gain | AI | is given by—
     	36 × 103

1. 1. 400

   
   
   

   2. 25

   
   
   

   3. 90

   
   
   

   4. 100

   
   
   

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   In the common amplifier, the current gain is given by
   

   |AI | =		–hfe	=	100	100
   		1 + hoe · RL		1 + (4000 / 36 × 103)	1 + (1 / 9)

   
   or |A_I | = 90
   Hence alternative (C) is the correct answer.

   Correct Option: C

   In the common amplifier, the current gain is given by
   

   |AI | =		–hfe	=	100	100
   		1 + hoe · RL		1 + (4000 / 36 × 103)	1 + (1 / 9)

   
   or |A_I | = 90
   Hence alternative (C) is the correct answer.

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415. In the potential divider bias to obtain base current, R_B is the equivalent base resistor and R_E the emitter resistor—

1. 1. R_B > > R_E improves S₁

   
   
   

   2. R_B >> R_E improves S_B

   
   
   

   3. R_B << R_E improves both S₁ and S_B

   
   
   

   4. R_B has no effect on the stabilisation factors

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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