1. For the switching converter shown in the following figure, assume steady-state operation. Also assume that the components are ideal, the inductor current is always positive and continuous and switching period is T₅. If the voltage V_L is as shown, the duty cycle of the switch S is ________.
   

1. 1. 75

   
   
   

   2. 7.5

   
   
   

   3. 0.75

   
   
   

   4. 0.075

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_S = 15V
   V₀ = V_S = - 45
   V₀ = V_V + 45
   = (15 + 45) V = 60V
   

   Now, V0 =	VS	= %
   	1 - D

   
   

   ∴ 60 =	15	= %
   	1 - D

   
   ∴ 60(1 – D) = 15
   

   ∴ 1 - D =	1
   	4

   
   

   ∴ D =	3	= 0.75
   	4

   Correct Option: C

   V_S = 15V
   V₀ = V_S = - 45
   V₀ = V_V + 45
   = (15 + 45) V = 60V
   

   Now, V0 =	VS	= %
   	1 - D

   
   

   ∴ 60 =	15	= %
   	1 - D

   
   ∴ 60(1 – D) = 15
   

   ∴ 1 - D =	1
   	4

   
   

   ∴ D =	3	= 0.75
   	4

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2. A symmetrical square wave of 50% duty cycle has amplitude of ± 15V and time period of 0.4π ms. This square wave is applied across a series RLC circuit with R = 5Ω, L = 10 mH, and C = 4μF. The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is _________.
   

1. 1. 190 - 192

   
   
   

   2. 193 - 195

   
   
   

   3. 196 - 198

   
   
   

   4. 198 - 199

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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3. The circuit shown in meant to supply a resistive load R_L from two separate DC voltage sources. The switches S₁ and S₂ are controlled so that only one of them is ON at any instant. S₁ is turned on for 0.2 ms and S₂ is turned on for 0.3 ms in a 0.5 ms switching cycle time period. Assuming continuous conduction of the inductor current and negligible ripple on the capacitor voltage, the output voltage Vo (inVolt) across RL is ________.
   

1. 1. 3

   
   
   

   2. 5

   
   
   

   3. 7

   
   
   

   4. 9

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   V0 =	10 × 0.2 + 5 × 0.3	= 7 V
   	0.5

   Correct Option: C

   
   

   V0 =	10 × 0.2 + 5 × 0.3	= 7 V
   	0.5

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4. The ac bridge shown in the figure given below is used to measure the impedance Z.
   If the bridge is balanced for oscillator frequency f = 2 kHz, then impedance Z will be
   

1. 1. (260 + j0)Ω

   
   
   

   2. (0 + j200)Ω

   
   
   

   3. (260 – j200)Ω

   
   
   

   4. (260 + j200)Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   X_L = 2πfL
   = 2π × 2000 × 15.91 × 10^–3
   
   

   XC =	1
   	2πfC

   
   

   =	1	= 200Ω
   	2π × 2000 × 0.398 × 10-6

   
   Under balanced condition
   Z₁ Z₄ = Z₂ Z₃
   ⇒ R₁ (R₄ + jX₄) = (R₂ + jX₂) (R₃ – jX₃)
   ⇒ 500(R₄ + jX₄) = (300 + j200) (300 – j200)
   ⇒ 500(R₄ + jX₄) = 90000 + 40000
   

   ⇒ R4 + jR4 =	130000	= 260
   	500

   
   ⇒ Z = 260 + j0
   Alternately
   

   At balance	ZAB	=	ZBC
   	ZAD		ZCD

   
   

   or,	500	=	R + (1/jωc)
   	R + jωL		Z

   
   

   or Z =	(R + jωL)(R + 1/jωc)	= (300 + j2 × 10³ × 3.14 × 2 × 15.91 × 10-3)
   	500

   
   

   = 300 -	j/2 × 3.14 × 103 × 2 × 0.398 × 10-6	= %
   	500

   
   

   =	(300 + j199.82)(300 - j200)
   	500

   
   ≈ (260 + j0)Ω

   Correct Option: A

   X_L = 2πfL
   = 2π × 2000 × 15.91 × 10^–3
   
   

   XC =	1
   	2πfC

   
   

   =	1	= 200Ω
   	2π × 2000 × 0.398 × 10-6

   
   Under balanced condition
   Z₁ Z₄ = Z₂ Z₃
   ⇒ R₁ (R₄ + jX₄) = (R₂ + jX₂) (R₃ – jX₃)
   ⇒ 500(R₄ + jX₄) = (300 + j200) (300 – j200)
   ⇒ 500(R₄ + jX₄) = 90000 + 40000
   

   ⇒ R4 + jR4 =	130000	= 260
   	500

   
   ⇒ Z = 260 + j0
   Alternately
   

   At balance	ZAB	=	ZBC
   	ZAD		ZCD

   
   

   or,	500	=	R + (1/jωc)
   	R + jωL		Z

   
   

   or Z =	(R + jωL)(R + 1/jωc)	= (300 + j2 × 10³ × 3.14 × 2 × 15.91 × 10-3)
   	500

   
   

   = 300 -	j/2 × 3.14 × 103 × 2 × 0.398 × 10-6	= %
   	500

   
   

   =	(300 + j199.82)(300 - j200)
   	500

   
   ≈ (260 + j0)Ω

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5. Match the following:
   S. Electrodynamometer

   Instrument Type	Used for
   P. Permanent magnet moving coil	1. DC only
   Q. Moving iron connected through current transformer	2. AC only
   R. Rectifier	3. AC and DC

   
   

1. 1. P Q R S
      1 2 1 3

   
   
   

   2. P Q R S
      1 3 1 2

   
   
   

   3. P Q R S
      1 2 3 3

   
   
   

   4. P Q R S
      3 1 2 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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