76. A fully controlled converter bridge feeds a highly inductive load with ripple free load current. The input supply (v_s) to the bridge is a sinusoidal source. Triggering angle of the bridge converter is α = 30°. The input power factor of the bridge is ___.
    
    

1. 1. 0.74 to 0.82

   
   
   

   2. 82

   
   
   

   3. 7.4

   
   
   

   4. 74

   
   
   

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1. View Hint View Answer Discuss in Forum

   P_in = V _s,rms I _s,rms cos φ = V₀ I₀
   

   ⇒ cos φ =	V0 I0	[Since I0 = I s,rms ]
   	V s,rms I s,rms

   
   

   =	V0
   	V s,rms

   
   

   cos φ =		2 √2 Vs	cos30°		= 0.779
   		π
   		Vs

   
   

   Correct Option: A

   P_in = V _s,rms I _s,rms cos φ = V₀ I₀
   

   ⇒ cos φ =	V0 I0	[Since I0 = I s,rms ]
   	V s,rms I s,rms

   
   

   =	V0
   	V s,rms

   
   

   cos φ =		2 √2 Vs	cos30°		= 0.779
   		π
   		Vs

   
   

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77. The figure shows one period of the output voltage of an inverter. α should be chosen such that 60° < α < 90°. If rms value of the fundamental component is 50V, then α in degree is __________.
    

1. 1. 61-65

   
   
   

   2. 76-77

   
   
   

   3. 36-47

   
   
   

   4. All of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   =	4Vs	[1 - cosα - cosα + 0]
   	π

   
   

   =	4Vs	[1 - 2 cosα]
   	π

   
   

   RMS Value of V01 =	4Vs	[1 - 2 cosα]
   	π √2

   
   

   ⇒ 50 =	400	[1 - 2 cosα]
   	π √2

   
   α = 77.15°
   

   Correct Option: B

   
   

   =	4Vs	[1 - cosα - cosα + 0]
   	π

   
   

   =	4Vs	[1 - 2 cosα]
   	π

   
   

   RMS Value of V01 =	4Vs	[1 - 2 cosα]
   	π √2

   
   

   ⇒ 50 =	400	[1 - 2 cosα]
   	π √2

   
   α = 77.15°
   

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78. Figure (i) shows the circuit diagram of a chopper. The switch S in the circuit in Figure (i) is switched such that the voltage V_D across the diode has the wave shape as shown in Figure (ii). The capacitance C is large so that the voltage across it is constant. If switch S and the diode are ideal, the peak to peak ripple (in A) in the inductor current is __________
    
    

1. 1. 0.25

   
   
   

   2. 25

   
   
   

   3. 2.5

   
   
   

   4. 12.5

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Peak to peak ripple in inductor current in
   

   ∆I =	Vs - Vo	Tm
   	L

   
   

   =	100 -		100 ×	0.05		× 0.05 × 10-3
   				0.1
   			1 × 10-3

   
   = 50 × 0.05
   ⇒ ∆I = 2.5 A
   

   Correct Option: C

   
   Peak to peak ripple in inductor current in
   

   ∆I =	Vs - Vo	Tm
   	L

   
   

   =	100 -		100 ×	0.05		× 0.05 × 10-3
   				0.1
   			1 × 10-3

   
   = 50 × 0.05
   ⇒ ∆I = 2.5 A
   

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79. The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10 Ω and an inductance 0.05H connected in series. Assuming ideal thyristor and ideal diode, the thyristor firing angle (in degree) needed to obtain an average load voltage of 70V is ____________.
    

1. 1. 69.3

   
   
   

   2. 39.39

   
   
   

   3. 19.93

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   V₀ = 70v; Load, R = 10Ω ; L = 0.05H
   Firing angled = ?
   The o/p volt of given converter is
   

   V0 =	Vm	[1 + cosα]
   	2π

   
   

   70 =	525	[1 + cosα]
   	2π

   
   α = 69.3°
   
   

   Correct Option: A

   
   V₀ = 70v; Load, R = 10Ω ; L = 0.05H
   Firing angled = ?
   The o/p volt of given converter is
   

   V0 =	Vm	[1 + cosα]
   	2π

   
   

   70 =	525	[1 + cosα]
   	2π

   
   α = 69.3°
   
   

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80. In the given rectifier, the delay angle of the thyristor T₁ measured from the positive going zero crossing of V_s is 30°. If the input voltage V_s is 100 sin(100πt)V, the average voltage across R (in Volt) under steady-state is _____________.
    

1. 1. 11.25

   
   
   

   2. 61.52

   
   
   

   3. 21.52

   
   
   

   4. 71.25

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : α = 30° ,
   V_in = 100 sin(l00πt)
   

   Now , V0 =	Vm	[3 + cosα]
   	2π

   
   

   =	100	[3 + cos30°] = 61.52 V
   	2π

   Correct Option: B

   Given : α = 30° ,
   V_in = 100 sin(l00πt)
   

   Now , V0 =	Vm	[3 + cosα]
   	2π

   
   

   =	100	[3 + cos30°] = 61.52 V
   	2π

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