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Physics NEET MCQ
Quiz 6
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Q.1
At resonance, the source current is : [ Raj. 1996]
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a) minimum in series L-C-R circuit
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b) minimum in a parallel L-C-R circuit
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c) maximum in both series and parallel L-C-R circuits
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d) minimum in both series and parallel L-C-R circuits
Explanation
Answer:(b)
Q.2
In an A.C circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is 0.5 henry and of capacitance is 8µF. The angular frequency of the input A.C. voltage must be equal to [ CPMT 1991]
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a) 500
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b) 5×104
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c) 4000
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d) 5000
Explanation
use formula angular frequency = 1/ √LC Answer: (a)
Q.3
A resistance R = 12Ω, inductance(L) = 2 henry and capacitance (C) = 5 mF are connected un series to an a.c. generator of frequency 50Hz [ CPMT 1998]
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a) At resonance, the circuit impedance is zero
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b) At resonance, the circuit impedance is 12Ω
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c) The resonance frequency of the circuit is (1/2)π
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d) The inductive reactance is less than the capacitance
Explanation
Use formula Z = √(R2 + (XL - XC)2 Z= 12Ω Answer: (b)
Q.4
An LCR circuit contains resistance of 100Ω and supply of 200 volts at 300 rad/sec angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60°. If on the other hand only inductor is taken out, the current leads by 60° with the applied voltage. The current flowing in the circuit is [ CPMT 1998]
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a) 1 A
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b) 1.5A
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c) 2.0A
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d) 2.5A
Explanation
Circuit contains inductance : tan60 = ωL / R Circuit contains capacitor tan60 = 1 / ωCR Thus ωL / R = 1 / ωCR ωL = 1 /ωC is the condition for resonance At resonance circuit becomes resistive The current in the circuit irms = Erms/R = 200/100 = 2 amp. Answer:(c)
Q.5
An a.c circuit consists of an inductor of inductance 0.5H and a capacitor of capacitance 8µF in series. The current in the circuit is maximum when the angular frequency of a.c. source is [ CPMT 1986]
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a) 500 Hz
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b) 2×105
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c) 4000 Hz
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d) 5000 Hz
Explanation
Current is maximum at resonance frequency F = 1/√(LC) Answer: (a)
Q.6
An alternating current circuit consists of an inductance and a resistance in series. In this circuit [ CPMT 1986 ]
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a) The potential difference across and current in resistance leads the potential difference across inductance
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b) The potential difference across and current in resistance lags behind the potential difference across inductance by an angle π/2
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c) The potential difference across and current in resistance lags behind the potential difference across inductance by an angle π
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d) The potential difference across resistance lags behind the potential difference across inductance by an angle π/2
Explanation
Answer: (d)
Q.7
The average power dissipation in pure capacitance in A.C. us [ MPPMT 1998]
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a) ½ CV2
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b) CV2
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c) ¼ CV2
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d) zero
Explanation
Answer: (d)
Q.8
In non-resonant circuit, if the frequency is greater than the resonant frequency, the nature of circuit will be [ Raj PMT 1996]
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a) Resistive
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b) Capacitive
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c) Inductive
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d) All the above
Explanation
If we increase the frequency above the resonant frequency, XC (= 1/ωC) decreases and XL (= ωL) increases. If we increase the frequency far enough, XC becomes much smaller than XL, and the contribution from the LC part of the circuit becomes mostly inductive. If we lower the frequency below the resonant frequency, XC (= 1/ωC) increases and XL (= ωL) decreases. If we lower the frequency far enough, XC becomes much larger than XL, and the contribution from the LC part of the circuit becomes mostly capacitive. Answer:(b)
Q.9
The power factor in circuit connected to an A.C. power supply has a value which is [ MPPMT 1987]
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a) unity when the circuit contains an ideal inductance only
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b) unity when the circuit contains an ideal resistance only
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c) Zero when the circuit contains an ideal resistance only
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d) Unity when the circuit contains an ideal capacitance only
Explanation
Answer: (b)
Q.10
An LCR circuit containing R = 50Ω, L=1mH and C= 0.1µF. The impedance of the circuit will be minimum for a frequency of [ CET 1995]
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a) 105 / 2π s-1
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b) 106 / 2π s-1
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c) 2π×10-5 s-1
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d) 2π×10-6 s-1
Explanation
Impedance will be minimum at resonant frequency use formula ω = 1/√(LC) Answer: (a)
Q.11
In L-C-R series A.C. circuit, the phase angle between current and voltage is [ MPPMT 1998]
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a) Any angle between 0 and ± π/2
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b) π/2
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c) π
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d) Any angle between 0 and π/2
Explanation
Answer: (a)
Q.12
In pure inductive circuit, the current: [ MPPMT 1993]
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a) lags behind the applied emf by an angle π
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b) lags behind the applied emf by an angle π/2
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c) leads the applied emf by an angle π/2
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d) and applied emf are in same phase
Explanation
Answer:(b)
Q.13
Using an AC voltmeter, the potential difference in the electrical line in a house is read to be 234 Volts. IF the line frequency is known to be 50 cycles per second, the equation for the line voltage is [ MPPMT 1987]
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a) V = 165sin(100πt)
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b) V = 331sin(100πt)
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c) V = 234sin(100πt)
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d) V = 440sin(100πt)
Explanation
Answer: (b)
Q.14
The average power dissipation in a pure capacitor in AC circuit is [ MPMT 1998]
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a) ½ CV2
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b) CV2
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c) 2CV2
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d) zero
Explanation
Answer: (d)
Q.15
In an A.C circuit with voltage V and current I, the power dissipated is [ CBSE 1997]
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a) VI
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b) VI/8
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c) VI / √
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d) depends on the phase difference between I and V
Explanation
Power P = VrmsIrmscosθ here θ is phase difference between I and V Answer: (d)
Q.16
A choke is used in fluorescent tube to [ Raj.PMT 1996]
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a) increase current
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b) decrease current
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c) increase voltage momentary
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d) decrease voltage momentary
Explanation
Answer:(b)
Q.17
An AC source is connected to a resistive circuit. What is true of the following? [ CPMT 1985]
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a) current leads ahead of voltage in phase
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b) current lag behind of voltage in phase
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c) current and voltage are in same phase
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d) any of the above may be true depending upon the value of resistance
Explanation
Answer: (c)
Q.18
In the LCR series circuit the voltmeter and ammeter readings are :
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a) V = 100 Volts, I = 2 amp
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b) V = 100 volts, I = 5 amp
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c) V = 100 volts, I = 3 amp
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d) V = 100 volts, I = 1 amp
Explanation
V2 = VR2 + (VL2 - VC2 100 = VR VR = IR 100 = I×50 I = 2 amp Answer: (a)
Q.19
The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25mThe coil is rotated at the rate of 100 cycles per sec in a magnetic field of 0.2 weber/mThe peak value of the emf generated is nearly [ AMU 1995]
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a) 786kV
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b) 440kV
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c) 220kV
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d) 157.1 kV
Explanation
Vmax = NABω Answer: (d)
Q.20
A 20 volt A.C is applied to a circuit consisting of resistance and a coil with negligible resistance. If the voltage across the rsistance is 12Volts, the voltage across the coil is [ Raj. PMT 1997]
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a) 16 volts
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b) 10 volts
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c) 8 volts
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d) 6 volts
Explanation
V2 = VR2 + VL2 Answer: (a)
Q.21
A generator produces a voltage that is given by V = 240sin(120 t) volt, where t is in second. The frequency and rms voltage are [ MPPMT 1993]
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a) 60 Hz and 240 Volts
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b) 19 Hz and 120 Volts
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c) 19 Hz and 170 Volts
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d) 754 Hz and 170 Volts
Explanation
Vrms = Vo / √2 ω = 100 &2πf = 100 Answer:(c)
Q.22
An AC generator produced an output voltage. E = 170sin(377 t) volts, where t is in seconds. The frequency of AC voltage is [ MPPMT 1994]
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a) 50Hz
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b) 110Hz
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c) 60HZ
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d) 230HZ
Explanation
Answer: (c)
Q.23
The number of turns in the primary and the secondary of a transformer are 1000 and 3000 respectively. If 80 volt A.C. is applied to the primary coil of the transformer, then the potential difference per turn of the secondary coil would be [ CPMT 1990]
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a) 240 Volts
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b) 2400 volts
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c) 0.24 volts
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d) 0.08volts
Explanation
Answer:(d)
Q.24
In LCR circuit having L = 8.0 henry, C= 0.5µF and R = 100Ω in series. the resonance frequency ( in per second) is [ CPMT 1990]
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a) 600 radian/sec
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b) 600 hertz
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c) 500 radian/sec
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d) 500 hertz
Explanation
Answer: (c)
Q.25
An alternating voltage is connected in series with resuistance R and an inductance L. IF the voltage across the resistance is 200 volts and across the inductance is 150 volts the applied voltage is [ CPMT 1990]
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a) 350 vol
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b) 250 volt
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c) 500 volt
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d) 300 volt
Explanation
Answer: (b)
Q.26
A coil having an inductance of 1/π H is connected in series with a resistance of 300Ω. If 20V, 200Hz, AC source is impressed across the combination, the phase angle between voltage and current is [ JIPMER 1998]
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a) tan⁻¹(5/4)
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b) tan⁻¹(4/5)
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c) tan⁻¹(3/4)
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d) tan⁻¹(4/3)
Explanation
tanφ = Lω / R Answer: (d)
Q.27
In an L-C circuit the phase difference between current I and voltage V is [ CPMT 1999]
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a) π/2
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b) 0
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c) -π/2
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d) π
Explanation
Answer:(c)
Q.28
In a series circuit: R = 300Ω, L = 0.9H, C=2.0µF and ω = 1000rad/s The impedance of the circuit is [ PMT 1995]
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a) 1300Ω
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b) 900Ω
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c) 500Ω
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d) 400Ω
Explanation
Answer: (c)
Q.29
Hot wire ammeter are used for measuring ..[ Raj.PMT 1997]
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a) d.c.only
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b) a.c. only
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c) neither d.c nor a.c.
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d) both a.c. and d.c.
Explanation
Answer: (d)
Q.30
For series LCR circuit, wrong statement is [ Raj.PMT 1997]
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a) Applied emf and potential difference across resistance are in same phase
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b) Applied emf and potential difference at inductor coil have phase difference of π/2
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c) Potential difference at capacitor and inductor have phase difference of π/2
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d) phase difference between the potential difference between resistance and capacitor is π/2
Explanation
Answer: (c)
0 h : 0 m : 1 s
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