131. The figure given below shows a three- phase delta connected load supplied from a 400V, 50Hz, 3-phase balanced source. The pressure coil (PC) and cur r ent coi l (CC) of a wat t met er ar e connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection.
     The wattmeter reading will be
     
     

1. 1. 0

   
   
   

   2. 1600 Watt

   
   
   

   3. 800 Watt

   
   
   

   4. 400 Watt

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Assume Vab as reference phasor, we have
   V_ab = V ∠0°
   V_bc = 400 ∠– 120°
   V_ca = 400 ∠– 240°
   Current through current coil,
   

   Icc =	Vca	=	400 ∠– 240°
   	Z2		100 + 10

   
   = 4 ∠– 240° A
   and, voltage across pressure coil,
   V_pc = V_bc = 400∠– 120°
   Hence, wattmeter reading, P = V_pc I_cc cos φ
   φ is the angle between V_pc and I_cc
   = 400 × 4 × cos 120° = – 800 watts
   

   Correct Option: C

   
   Assume Vab as reference phasor, we have
   V_ab = V ∠0°
   V_bc = 400 ∠– 120°
   V_ca = 400 ∠– 240°
   Current through current coil,
   

   Icc =	Vca	=	400 ∠– 240°
   	Z2		100 + 10

   
   = 4 ∠– 240° A
   and, voltage across pressure coil,
   V_pc = V_bc = 400∠– 120°
   Hence, wattmeter reading, P = V_pc I_cc cos φ
   φ is the angle between V_pc and I_cc
   = 400 × 4 × cos 120° = – 800 watts
   

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132. An average-reading digital multimeter reads 10 V when fed with a triangular wave, symmetric about the time-axis. For the same input an rmsreading meter will read
     

1. 1. 	20
      	√3

      
      

   
   
   

   2. 	10
      	√30

      
      

   
   
   

   3. 20 √30
      

   
   
   

   4. 10 √30
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   VAV =	1	× Vmax ×	T
   	2		2	= 10
   		T
   		2

   
   ∴ V_max = 10 × 2 = 20V
   Equation of line OA, υ = mt
   

   At t =	T	, υ = 20
   	4

   
   

   ∴ m =	20	=	80
   	T / 4		T

   
   

   ⇒ υ =	80	t
   	T

   
   
   

   Correct Option: A

   VAV =	1	× Vmax ×	T
   	2		2	= 10
   		T
   		2

   
   ∴ V_max = 10 × 2 = 20V
   Equation of line OA, υ = mt
   

   At t =	T	, υ = 20
   	4

   
   

   ∴ m =	20	=	80
   	T / 4		T

   
   

   ⇒ υ =	80	t
   	T

   
   
   

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133. The Thevenin’s equivalent of a circuit operating at ω = 5 rad/s, has V_oc = 3.71 ∠ – 15.9° V and Z₀ = 2.38 – j 0.667 Ω . At this frequency, the minimal realization of the Thevenin’s impedance will have a
     

1. 1. resistor and a capacitor and an inductor

   
   
   

   2. resistor and a capacitor

   
   
   

   3. resistor and an inductor

   
   
   

   4. capacitor and an inductor

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since Thevenin’s impedence,
   Z₀ = 2.38 – j 0.667 Ω
   Z₀ is a combination of resistor and capacitor in series
   For Minimum realization,
   

   Z0 = R +	1	= R +	1	= R -	j
   	Cs		jωC		ωC

   
   So Z₀ will have a resistor and a capacitor.

   Correct Option: B

   Since Thevenin’s impedence,
   Z₀ = 2.38 – j 0.667 Ω
   Z₀ is a combination of resistor and capacitor in series
   For Minimum realization,
   

   Z0 = R +	1	= R +	1	= R -	j
   	Cs		jωC		ωC

   
   So Z₀ will have a resistor and a capacitor.

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134. The time constant for the given circuit will be
     
     

1. 1. 	1	s
      	9

      
      

   
   
   

   2. 	1	s
      	4

      
      

   
   
   

   3. 4 s
      

   
   
   

   4. 9 s

   
   
   

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1. View Hint View Answer Discuss in Forum

   For time constant
   
   

   ∴ Time constant = Req Ceq = 6 ×	2	= 4 sec
   	3

   Correct Option: C

   For time constant
   
   

   ∴ Time constant = Req Ceq = 6 ×	2	= 4 sec
   	3

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135. The resonant frequency for the given circuit will be
     
     

1. 1. 1 rad/s

   
   
   

   2. 2 rad/s

   
   
   

   3. 3rad/s

   
   
   

   4. 4 rad/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Let frequency be ω .
   

   ∴ z = jωL +	R
   	1 + jωRC

   
   

   z = j 0.1 ω +	1	×	1 - jω
   	1 + jω		1 - jω

   
   

   = j0.1ω +	1 - jω
   	1 + ω2

   
   

   =	1	+ j		0.1 ω -	ω
   	1 + ω2				1 + ω2

   
   For resonance, imaginary part must be zero.
   

   ∴ 0.1 -	ω	= 0
   	1 + ω2

   
   

   or 0.1 =	1
   	1 + ω2

   
   or 1 + ω² = 10
   or ω² = 9
   ω = 3 rad/sec
   

   Correct Option: C

   
   Let frequency be ω .
   

   ∴ z = jωL +	R
   	1 + jωRC

   
   

   z = j 0.1 ω +	1	×	1 - jω
   	1 + jω		1 - jω

   
   

   = j0.1ω +	1 - jω
   	1 + ω2

   
   

   =	1	+ j		0.1 ω -	ω
   	1 + ω2				1 + ω2

   
   For resonance, imaginary part must be zero.
   

   ∴ 0.1 -	ω	= 0
   	1 + ω2

   
   

   or 0.1 =	1
   	1 + ω2

   
   or 1 + ω² = 10
   or ω² = 9
   ω = 3 rad/sec
   

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