Important AC Circuit MCQs
In this post, some of the AC Circuit MCQs are given. It includes upper cut off frequency, half power, lower cut off frequency, power factor at half power points, upper cut off frequency at resonance, lower cut off frequency at resonance, Q factor of series resonance circuit, Q factor is ratio of, high Q factor of series resonance circuit, meaning of higher Q factor, power dissipation at series resonance frequency and power dissipation at off series resonance frequency.
AC Circuit MCQs: 181 – 185
181 The upper cut off angular frequency at half power is given by
( a )ωo ( 1 – 1/2Qo )
( b )ωo ( 1 + 1/2Qo )
( c )ωo × Qo
( d )ωo / Qo
Correct Answer ( b ): ωo( 1 + 1/2Qo )
182 The lower cut off angular frequency at half power is given by
( a )ωo × Qo
( b )ωo / Qo
( c )ωo( 1 – 1 / 2 Qo )
( d )ωo ( 1 + 1 / 2Qo )
Correct Answer ( c ): ωo( 1 – 1/2 Qo )
183 Which of the following relation is true at half power points?
( a )Ro = Xo
( b )Ro = ± Xo
( c )Xo= Zo
( d )None of the above
Correct Answer ( b ): Ro = ± Xo
184 The power factor at half power point is equal to
( a )Unity
( b )0.707
( c )Zero
( d )None of the above
Correct Answer ( b ): 0.707
As Z = R at the resonance frequency therefore power factor at the resonance frequency is unity and current is maximum
At half power points, resistance is equal to reactance R = X
Therefore Z = √ R2 + R2 = √2R
Cos Φ = R/Z = R/√2R = 1/√2
Therefore Φ = 0.707 leading at point A ( less than resonance frequency ) and lagging at point B ( greater than resonance frequency )
185 The lower cut off frequency at resonance condition is
( a )fo – ( R / 4πL )
( b )fo + ( R / 4πL )
( c )fo – ( R / 2πL )
( d )fo+ ( R / 2πL )
Correct Answer ( a ): fo – ( R / 4πL )
AC Circuit MCQs: 186 – 190
186 The upper cut off frequency at resonance is given by______. (where fo is centre frequency )
( a )fo + ( R / 2πL )
( b )fo + ( R / 4πL )
( c )fo – ( R / 2πL )
( d )fo – ( R / 4πL )
Correct Answer ( b ): fo + ( R / 4πL )
187 Which of the following relation is true at resonance condition? (Where fo is centre frequency)
( a )Δf = fo × Qo
( b )Δf = Qo / fo
( c )Δf = fo / Qo
( d )Δf = f1 + f2
Correct Answer ( c ): Δf = fo / Qo
188 The Q Factor of the series resonance circuit is related to
( a )Voltage magnification
( b )Current magnification
( c )Power magnification
( d )All of the above
Correct Answer ( a ): Voltage magnification
Higher value of Q factor of the resonance curve means higher selectivity of the coil or higher voltage magnification.
189 The Q factor of a series resonance circuit is
( a )1 / ωoRC
( b )1 / ωoRL
( c )ωo / RL
( d )ωo / RC
Correct Answer ( a ): 1 / ωoRC
Q factor Qo = ωo / Bandwidth
= ωoL / R
= 1 / ωoCR
= 1/R √( L/C)
= fo / f2 – f1
190 The Q factor of a series resonance circuit is ratio of
( a )Active power / apparent power
( b )Active power / reactive power
( c )Reactive power / active power
( d )None of the above
Correct Answer ( c ): Reactive power / active power
Q factor of series resonance = Reactive power/ Active power
= IoXLo / IoR
= XLo / R
= ωoL / R
OR
Q factor Qo = ωo / Bandwidth
= ωoL / R
= 1 / ωoCR
= 1/R √( L/C)
= fo / f2 – f1
AC Circuit MCQs: 191 – 197
191 The Q factor of a series resonance circuit is ______. (Where Ф is power factor of the coil)
( a )Sin Ф
( b )Cos Ф
( c )tan Ф
( d )None of the above
Correct Answer ( c ): tan Ф
Q factor of series resonance circuit = ωoL / R = tan Φo
192 The quality factor of a series resonant circuit is given by
( a )√ ( L / C )
( b )( 1 / R ) √ ( L / C )
( c )R √ ( L / C )
( d )R / √ ( LC )
Correct Answer ( b ): ( 1 / R ) √ ( L / C )
Q factor of series resonance = ωoL / R
Qo = 2πfoL/R
Now, resonance frequency fo = 1/2π√ (LC)
2πfo = 1/ √ (LC)
Therefore Q factor = 1{ √ (L/C) } / R
193 The high Q factor of a series resonance circuit means higher
( a )Current magnification
( b )Voltage magnification
( c )Selectivity
( d )Both ( b ) and ( c )
Correct Answer ( d ): Both ( b ) and ( c )
194 Higher value of Q factor means______ value of resistance of coil.
( a )High
( b )Medium
( c )Small
( d )Zero
Correct Answer ( c ): Small
Q factor = ωoL / R
High value of Q factor means lower value of resistance of higher value of inductance
195 The Q factor of series resonance circuit is
( a )fo / ( f2 – f1 )
( b )fo / ( f2 × f1 )
( c )( f2 – f1 ) / fo
( d )√ ( fo / f2 × f1 )
Correct Answer ( a ): fo / ( f2 – f1)
Q factor Qo = ωo / Bandwidth
= 2π fo / 2π f2 – 2π f1
= fo / ( f2 – f1)
196 The power dissipation in the series resonance circuit becomes maximum at
( a )XL > XC
( b )XL < XC
( c )XL = XC
( d )XC = 0
Correct Answer ( c ): XL = XC
At series resonance frequency, Z = R or ( XL = XC ) therefore current value become maximum at resonance frequency so power dissipation maximum at series resonance.
197 The power dissipation at off resonance frequency is reduced by a factor ______. (Where Q is tangent of the circuit phase angle)
( a )1 / ( 1 + Q )
( b )1 / ( 1 − Q )
( c )1 / ( 1 + Q2)
( d )1 / ( 1 − Q2)
Correct Answer ( c ): 1 / ( 1 + Q2 )
Power at resonance frequency Po = V2 / R
Power at off resonance frequency, frequency other than resonance frequency
P = I2R
= (V/Z)2R
= V2R / Z2
= V2R / R2 + X2
= V2R / R2 + X2R2/R2
= V2R / R2 + Q2R2 ( Where Q = tangent of circuit phase angle = tan θ
= V2R / R2 ( 1 + Q2 )
= V2 / R ( 1 + Q2 )
= Po / ( 1 + Q2 )
Therefore power at off resonance is equal to Po / ( 1 + Q2 )
Basic Electrical MCQs PDF
Basic Electrical MCQs PDF Download
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