Important MCQs of AC Circuits
In this post, some of the important MCQs of AC Circuits are given. It includes power loss in pure inductor, charge across capacitor, value of operator j, operator j2, operator j3, three phase operator a, operator a2, operator a3 and addition of three phase operators.
AC Circuit MCQs: 81 to 85
81 The average power loss in 50 µH pure inductance is
( a )50 watt
( b )500 watt
( c )100 watt
( d )Zero
Correct Answer ( d ): Zero
The power loss in the pure inductor is zero.
82 The charge across capacitor is given by
( a )IV
( b )CV
( c )I2R
( d )V2C
Correct Answer ( b ): CV
The charge across capacitor = CV = Q / V2 = Q )
( As Q = CV or C = Q / V )
83 The vector quantity is represented by
( a )Magnitude
( b )Direction
( c )Either ( a ) or ( b )
( d )Both ( a ) and ( b )
Correct Answer ( d ): Both ( a ) and ( b )
The vector is represented by both magnitude and direction whereas scaler is represented by magnitude only.
84 The scalar quantity is represented by
( a )Magnitude
( b )Direction
( c )Either ( a ) or ( b )
( d )Both ( a ) and ( b )
Correct Answer ( a ): Magnitude
The scaler is represented by magnitude only.
85 The symbol j represents anticlockwise rotation of vector quantity through
( a )0o
( b )45o
( c )90o
( d )180o
Correct Answer ( c ): 90o
The vector j is represented by anti-clockwise rotation of vector through 90 degree.
AC Circuit MCQs: 86 to 90
86 The value of operator j is
( a )1
( b )√ 1
( c )√ – 1
( d )– √ 1
Correct Answer ( c ): √ – 1
87 The operator j2 is equal to
( a )1
( b )√ 1
( c )– 1
( d )– √ 1
Correct Answer ( c ): – 1
As j = √ – 1, j2 = – 1
88 The operator j3 is equal to
( a )1
( b )√ 1
( c )– 1
( d )– j
Correct Answer ( d ): – j
As j = √ – 1
Now j3 = ( j2 )( j ) = – j ( because j2 is equal to – 1 )
89 The operator j4 is equal to
( a )1
( b )√ 1
( c )– 1
( d )√ – 1
Correct Answer ( a ): 1
As j = √ – 1 therefore j2 = – 1
Now j4 = j2 * j2 = (– 1)( – 1) = 1
90 The value of the 120o operator a in the three-phase circuit is
( a )1 + j 0 ( 1 + ∠90o)
( b )– 0.5 + j 0.866 ( 1 + ∠120o)
( c )– 0.5 – j 0.866 ( 1 + ∠-120o)
( d )– 1 + j 0 ( 1 + ∠180o)
Correct Answer ( b ): – 0.5 + j 0.866
The value of operator a rotates in the anti-clockwise direction
a = 1 ∠120o = 1 ( Cos 120o + j Sin 120o ) = – 0.5 + j 0.866
More information About Operator j
AC Circuit MCQs: 91 to 95
91 The value of the 120o operator a2 in the three-phase circuit rotates in the clockwise direction by
( a )120o
( b )240o
( c )– 120o
( d )0o
Correct Answer ( a ): 120o
The value of operator a rotates in the anti-clockwise direction, a2 means operated rotates 240 degree in the anticlockwise direction or 120 degree in the clockwise direction.
92 The value of three phase 120o operator a3 is
( a )0
( b )– 0.5
( c )0.5
( d )1
Correct Answer ( d ): 1
The operator a3 = 1 ∠360o = 1 ( Cos 360o + j Sin 360o ) = 1 + j 0 = 1
93 The value of three phase 120o operator a2 is
( a )– 0.5 + j 0.866
( b ) 0.5 + j 0.866
( c )– 0.5 – j 0.866
( d )None of the above
Correct Answer ( c ): – 0.5 – j 0.866
The operator a2 = 1 ∠240o = 1 ( Cos 240o + j Sin 240o ) = – 0.5 – j 0.866
94 Which of the following relation is true for the three-phase circuit operator a?
( a )a2 + a = – 1
( b )a2 + a = 0
( c )a2 + a = 1
( d )a2 + a = a
Correct Answer ( a ): a2+ a = – 1
We have to prove that a2+ a = – 1
Now
a2 = 1 ∠240o = 1 ( Cos 240o + j Sin 240o ) = – 0.5 – j 0.866 and
a = 1 ∠120o = 1 ( Cos 120o + j Sin 120o ) = – 0.5 + j 0.866
L.H.S. = a2+ a = ( – 0.5 – j 0.866 ) + ( – 0.5 + j 0.866 ) = – 1.0
95 Which of the following relation is true for the three phase 120o operator?
( a )a3 + a2 + a = 1
( b )a3+ a2 + a = ( – 1 )
( c )a3 + a2 + a = √ ( – 1 )
( d )a3 + a2 + a = 0
Correct Answer ( d ): a3+ a2 + a = 0
We have to prove that a3+ a2 + a = 0
Now
a3 = 1 ∠360o = 1 ( Cos 360o + j Sin 360o ) = 1 + j 0 = 1
a2 = 1 ∠240o = 1 ( Cos 240o + j Sin 240o ) = – 0.5 – j 0.866 and
a = 1 ∠120o = 1 ( Cos 120o + j Sin 120o ) = – 0.5 + j 0.866
L.H.S = a3+ a2 + a
= ( 1 ) + ( – 0.5 – j 0.866 ) + ( – 0.5 + j 0.866 ) = 0
AC Circuit MCQs: 96 to 100
96 The conjugate of ( – x + jy ) is
( a )( – x – jy )
( b )( – x + jy )
( c )( x – jy )
( d )( x + jy )
Correct Answer ( a ): ( – x – jy )
The conjugate of ( – x + jy ) is ( – x – jy )
97 The power factor of R – L series circuit is
( a )Unity
( b )Lagging
( c )Leading
( d )Any of the above
Correct Answer ( b ): Lagging
As the power factor of the inductor is lagging therefore the power factor of the R – L circuit is lagging.
98 The power loss in the R – L series circuit is due to
( a )Inductance
( b )Resistance
( c )Inductive reactance
( d )None of the above
Correct Answer ( b ): Resistance
There is no power loss in the pure inductive circuit therefore power loss in the RL circuit is only due to resistance of the circuit.
99 Which of the following is true for series RL circuit? VR = 3V, VL = 4V, then supply voltage V is
( a )7.0 Volt
( b )1.0 Volt
( c ).– 1.0 Volt
( d )5.0 Volt
Correct Answer ( d ): 5.0 Volt
The voltage vector in the RL Series OR RC Series circuit form Pythagoras therefore
V = √ VR2 + VL2 = √ (3)2 + (4)2 = √ 25 = 5.0 V
100 The power consumption in the series RL circuit
( a )VI
( b )VI Cos Ф
( c )VI Sin Ф
( d )Zero
Correct Answer ( b ): VI Cos Ф
The power consumption in the RL or RC or RLC circuit is equal to active power.
P = VI Cos Ф
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