1. The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 s is applied to the SCR. The maximum value of R in Ω to ensure successful firing of the SCR is _____.
   

1. 1. 7065

   
   
   

   2. 6045

   
   
   

   3. 6055 to 6065

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   From the given circuit,
   
   Where I_L is the latching current and t_g is the gate pulse width
   

   ⇒ 40 × 10-3 =	100	+	1		1 - e-{2.5 × 50 × 10-6} / 10-3}
   	R		5

   
   

   ⇒ 40 × 10-3 =	100	+	1		1 - e-0.125
   	R		5

   
   

   ⇒	100	= 0.01649
   	R

   
   ⇒ R = 6060 Ω.
   

   Correct Option: C

   From the given circuit,
   
   Where I_L is the latching current and t_g is the gate pulse width
   

   ⇒ 40 × 10-3 =	100	+	1		1 - e-{2.5 × 50 × 10-6} / 10-3}
   	R		5

   
   

   ⇒ 40 × 10-3 =	100	+	1		1 - e-0.125
   	R		5

   
   

   ⇒	100	= 0.01649
   	R

   
   ⇒ R = 6060 Ω.
   

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2. A step-up chopper is used to feed a load at 400 V dc fr om a 250 V dc source. The inductor current is continuous. If the ‘off’ time of the switch is 20 s, the switching frequency of the chopper in kHz is _____.

1. 1. 32.5

   
   
   

   2. 21.5

   
   
   

   3. 31.0 to 31.5

   
   
   

   4. 31.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_o = 400 V
   V_s = 250 V
   T_off = 20 μsec
   F = ?
   

   Vo =	Vs
   	1 - D

   
   

   ∴ 1 - D =	Vs
   	1 - D

   
   

   1 - D =	250
   	400

   
   Now T_off = (1 – D) T
   

   ∴ T =	Toff	=	20 × 10-6	× 400
   	1 - D		250

   
   

   Now f =	1	=	250 × 106
   	T		20 × 400

   
   ∴ f = 31.25 kHz
   

   Correct Option: C

   V_o = 400 V
   V_s = 250 V
   T_off = 20 μsec
   F = ?
   

   Vo =	Vs
   	1 - D

   
   

   ∴ 1 - D =	Vs
   	1 - D

   
   

   1 - D =	250
   	400

   
   Now T_off = (1 – D) T
   

   ∴ T =	Toff	=	20 × 10-6	× 400
   	1 - D		250

   
   

   Now f =	1	=	250 × 106
   	T		20 × 400

   
   ∴ f = 31.25 kHz
   

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3. A single-phase inverter is operated in PWM mode generating a single-pulse of width 2d in the centre of each half cycle as shown in the figure given below. It is found that the output voltage is free from 5th harmonic for pulse width 144°. What will be percentage of 3^rd harmonic present in the output voltage (V₀₃/V_{o1 max})?
   
   

1. 1. 0.0%
      

   
   
   

   2. 19.6%
      

   
   
   

   3. 31.7%
      

   
   
   

   4. 53.9%
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   The output of single-phase inverter,
   
   For pulse width of 144° , output is tree from
   5th harmonic i.e., 2d = 144°
   ⇒ d = 72°
   

   Then , V01 =	4Vs	sin 72° . sin ωt. sin	π
   	π		2

   
   

   V03 =	4Vs	sin(3 × 72°) . sin(3ωt) . sin		3π
   	3π			2

   
   

   ratio ,	V03	=			4Vs		.sin(216°)	≈ 20%
   					3π
   	V01 , max				4Vs		.sin(72°)
   					π

   
   

   Correct Option: B

   The output of single-phase inverter,
   
   For pulse width of 144° , output is tree from
   5th harmonic i.e., 2d = 144°
   ⇒ d = 72°
   

   Then , V01 =	4Vs	sin 72° . sin ωt. sin	π
   	π		2

   
   

   V03 =	4Vs	sin(3 × 72°) . sin(3ωt) . sin		3π
   	3π			2

   
   

   ratio ,	V03	=			4Vs		.sin(216°)	≈ 20%
   					3π
   	V01 , max				4Vs		.sin(72°)
   					π

   
   

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4. A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4 : 1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery (12 V dc) with the series current limiting resistor being 19.04 Ω. The charging current is
   

1. 1. 2.43 A

   
   
   

   2. 1.65 A

   
   
   

   3. 1.22 A

   
   
   

   4. 1.0 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   	V1	=	4	⇒ V2 =	V1	=	230	= 57.5 Volts
   	V2		1		4		4

   
   Diode will conduct when V₂ > E
   i.e., V_m sin θ₁ = E
   ⇒ (57.5) √2 sin θ = 12
   ⇒ θ₁ = 8.48° or, 0.148 rad
   
   

   =	1	[2Vm cosθ1 - E(π - 2θ1)]
   	2πR

   
   

   =	1	[2 × 57.5 √2 × cos 8.48° - 12(π - 2 × 0.148)] ≈ 1 A
   	2 π × 19.04

   Correct Option: A

   
   

   	V1	=	4	⇒ V2 =	V1	=	230	= 57.5 Volts
   	V2		1		4		4

   
   Diode will conduct when V₂ > E
   i.e., V_m sin θ₁ = E
   ⇒ (57.5) √2 sin θ = 12
   ⇒ θ₁ = 8.48° or, 0.148 rad
   
   

   =	1	[2Vm cosθ1 - E(π - 2θ1)]
   	2πR

   
   

   =	1	[2 × 57.5 √2 × cos 8.48° - 12(π - 2 × 0.148)] ≈ 1 A
   	2 π × 19.04

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5. A 220 V, 20 A, 1000 rpm, separately excited dc motor has an armature resistance of 2.5Ω. The motor is controlled by a step down chopper with a frequency of 1 kHz. The input dc voltage to the chopper is 250 V. The duty cycle of the chopper for the motor to operate at a speed of 600 rpm delivering the rated torque will be
   

1. 1. 0.518

   
   
   

   2. 0.608

   
   
   

   3. 0.852

   
   
   

   4. 0.902

   
   
   

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1. View Hint View Answer Discuss in Forum

   For rated speed,
   
   V = E + I _a R_a
   ⇒ E = V – I_a R_a = 220 – 20 × 2.5 = 170 V
   

   For 600 rpm, E2 =	E1	× N2
   	N1

   
   

   =	170	× 1600 = 102 V
   	1000

   
   Since T ∝ I
   therefore rated torque means rated current Now voltage applied,
   V_t = E₂ + I_a R_a = 120 + 20 × 2.5 = 152 V
   V₀ = δ V_in
   

   ⇒ δ =	V0	=	152	= 0.608
   	Vin		250

   
   Alternately
   V_t – I_a r_a = E_a
   220 – 20 × 2.5 = E_a
   E_a = 170 = Tω = T.2π(1000) ...(A)
   The output of chopper, V = α V_i = α 250 [α = duty cycle]
   The, 250 α – 20 × 2.5 = T.2π(600) ...(B)
   

   	170	=	1000
   	250 α - 50		600

   
   or, α = 0.608
   

   Correct Option: B

   For rated speed,
   
   V = E + I _a R_a
   ⇒ E = V – I_a R_a = 220 – 20 × 2.5 = 170 V
   

   For 600 rpm, E2 =	E1	× N2
   	N1

   
   

   =	170	× 1600 = 102 V
   	1000

   
   Since T ∝ I
   therefore rated torque means rated current Now voltage applied,
   V_t = E₂ + I_a R_a = 120 + 20 × 2.5 = 152 V
   V₀ = δ V_in
   

   ⇒ δ =	V0	=	152	= 0.608
   	Vin		250

   
   Alternately
   V_t – I_a r_a = E_a
   220 – 20 × 2.5 = E_a
   E_a = 170 = Tω = T.2π(1000) ...(A)
   The output of chopper, V = α V_i = α 250 [α = duty cycle]
   The, 250 α – 20 × 2.5 = T.2π(600) ...(B)
   

   	170	=	1000
   	250 α - 50		600

   
   or, α = 0.608
   

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