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NEET • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Mixed difficulty

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121 Medium
121 of 154
Analyzing +4 | -1
Given below are two statements:

Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

In the light of above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: C
Both Statement I and Statement II are false
Option C is correct because both statements about AC circuits and resonance are false. Statement I is incorrect as resonance requires both inductor and capacitor. Statement II is incorrect as pure capacitor or inductor circuits have a power factor of zero, hence consume no power.
122 Medium
122 of 154
Application +4 | -1
What will be the self inductance of a coil, to be connected in a series with a resistance of π3Ω\pi \sqrt 3 \mathrm{\Omega } such that the phase difference between the emf and the current at 50 Hz frequency is 30°
View answer and explanation
Correct option: D
0.01 henry
Option D (0.01 henry) is correct because the self-inductance is calculated using the tangent of the phase angle and the given resistance and frequency.
123 Medium
123 of 154
Understanding +4 | -1
In an ideal transformer the number of turns of primary and secondary coil is given as 100\mathrm{100} and 300\mathrm{300} respectively. If the power input is 60  W60\,\,W, the power output is
View answer and explanation
Correct option: D
60  W60\,\,W
Option D (60 W) is correct because in an ideal transformer, the power input equals the power output, assuming no losses.
124 Medium
124 of 154
Application +4 | -1
In a current I given by Iosin⁡[ωt−π2]{I_\mathrm{o}}\sin \left[ {\omega t - \frac{\pi }{2}} \right] flows in an AC circuit across which an AC potential of E=EoE = {E_\mathrm{o}} sin wt has been applied, then the power consumption PP in the circuit will be
View answer and explanation
Correct option: D
P = 0
Option D (P = 0) is correct because the phase difference between the voltage and current is π/2, resulting in zero average power consumption.
125 Medium
125 of 154
Remembering +4 | -1
The hysteresis cycle for the material of a transformer core is
View answer and explanation
Correct option: B
Tall and narrow
Option D (Short and narrow) is correct because a transformer core material should have low hysteresis loss, which is represented by a short and narrow hysteresis loop.
126 Medium
126 of 154
Analyzing +4 | -1
Power factor of an L-R series circuit is 0.6 and that of a C-R series circuit is 0.5 . If the elements ( L,CL, C and RR ) of the two circuit are joined in series the power factor of this circuit is found to be 1. The ratio of the resistance is L-R circuit to the resistance in the C-R circuit is
View answer and explanation
Correct option: D
334\frac{3 \sqrt{3}}{4}
Option D (334\frac{3 \sqrt{3}}{4}) is correct because it is the ratio of the resistances calculated from the given power factors and the condition of resonance.
127 Medium
127 of 154
Analyzing +4 | -1
Choose the incorrect statement
View answer and explanation
Correct option: D
When a current i=(1+2sin⁡ωt)i=(1+2 \sin \omega t) AA flows in a wire, d. c ammeter reads 2 A2 {~A}.
Option D (When a current i=(1+2sin⁡ωt)i=(1+2 \sin \omega t) AA flows in a wire, d. c ammeter reads 2 A2 {~A}.) is the incorrect statement because a DC ammeter reads the average value of the current, which is 1 A in this case, not 2 A.
128 Easy
128 of 154
Application +4 | -1
In an AC circuit, V and I are given by V=150sin⁡(150t)  volt  and  I=150sin⁡(150t+π3)ampV = 150\sin \left( {150\mathrm{t}} \right)\mathrm{\;volt\;and\;}I = 150\sin \left( {150\mathrm{t} + \frac{\mathrm{\pi }}{3}} \right)\mathrm{amp} . The power dissipated in the circuit is
View answer and explanation
Correct option: B
5625 W
Option B (5625 W) is correct because the power dissipated is given by P=VrmsIrmscos⁡ϕP = V_{rms}I_{rms}\cos\phi, where ϕ=π/3\phi = \pi/3, so P=(150/2)(150/2)cos⁡(π/3)=5625P = (150/\sqrt{2})(150/\sqrt{2})\cos(\pi/3) = 5625 W.
129 Easy
129 of 154
Understanding +4 | -1
A bulb is connected first with dc and then ac of same voltage it will shine brightly with
View answer and explanation
Correct option: D
Equally with both
Option D (Equally with both) is correct because the bulb's brightness depends on the RMS voltage, which is the same for both AC and DC in this scenario.
130 Medium
130 of 154
Application +4 | -1
A pure inductive coil of 30 mH is connected to an AC source of 220 V, 50 Hz. The rms current in the coil is
View answer and explanation
Correct option: B
23.4 A
Option B (23.4 A) is correct because the RMS current in an inductive circuit is calculated using the formula I=VXLI = \frac{V}{X_L}, where XL=ωL=2πfLX_L = \omega L = 2\pi fL.
131 Easy
131 of 154
Application +4 | -1
A transformer with turns ratio N1N2=501\frac{N_1}{N_2}=\frac{50}{1} is connected to a 120 volt AC supply. If primary and secondary circuit resistance are 1.5 kΩ1.5 \mathrm{~k} \Omega and 1 Ω1 ~\Omega respectively then find out power output.
View answer and explanation
Correct option: A
5.76 W5.76 \mathrm{~W}
Option A (5.76 W5.76 \mathrm{~W}) is correct because it is the power output calculated using the transformer turns ratio and the given resistances and voltage.
132 Easy
132 of 154
Application +4 | -1
An electric heater rated 220 V and 550 W is connected to A.C mains. The current drown by it is
View answer and explanation
Correct option: B
2.5 A
Option B (2.5 A) is correct because it is calculated using the formula I = P/V.
133 Hard
133 of 154
Understanding +4 | -1
Which of the following components of a L – C – R circuit, with AC supply, dissipates energy?
View answer and explanation
Correct option: B
R
Option B (R) is correct because the resistor (R) is the only component in an L-C-R circuit that dissipates energy in the form of heat due to its resistance to current flow.
134 Medium
134 of 154
Application +4 | -1
The primary winding of a transformer has 200 turns and its secondary winding has 50 turns. If the current in the secondary winding is 40 A, the current in the primary is
View answer and explanation
Correct option: A
10A
Option C (50 ohm) is correct because the impedance is calculated using the formula Z=R2+XL2Z = \sqrt{R^2 + X_L^2}, where XL=2πfLX_L = 2\pi fL.
135 Medium
135 of 154
Application +4 | -1
The values of L, C and R for a circuit are 1H, 9F and 3Ω. What is the quality factor for the circuit at resonance?
View answer and explanation
Correct option: C
19\frac{1}{9}
Option D (13\frac{1}{3}) is correct because the quality factor is calculated using the formula Q = 1/R * sqrt(L/C).
136 Medium
136 of 154
Analyzing +4 | -1
If L,CL,C and RR are the self-inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?
View answer and explanation
Correct option: D
LC\frac{L}{C}
Option D (LC\frac{L}{C}) is correct because the dimension of LC\frac{L}{C} is (Voltage/Current)2, which is not the dimension of time.
137 Medium
137 of 154
Application +4 | -1
If the direction of i is reversed, (VB−VA)\left( {{V_B} - {V_A}} \right) will be
View answer and explanation
Correct option: B
15 V
Option B is correct. Reversing the current direction changes the sign of the voltage drop across the resistor, leading to a new potential difference of 15 V.
138 Easy
138 of 154
Understanding +4 | -1
The instantaneous values of current and emf in an ac circuit are I=1/2I = 1/\sqrt 2 sin⁡314  t  amp\sin 314\;t\;amp and E=2sin⁡(314  t−π/6)VE = \sqrt 2 \sin \left( {314\;t - \pi /6} \right)V respectively. The phase difference between E and I will be
View answer and explanation
Correct option: A
−π/6  rad- \pi /6\;rad
Option A (−π/6  rad-\pi/6 \;rad) is correct because the phase difference is directly observed from the given equations of emf and current.
139 Medium
139 of 154
Application +4 | -1
In a series L-C-R circuit R=200 Ω and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the L-C-R circuit is
View answer and explanation
Correct option: D
242 W
Option D (242 W) is correct because at resonance, the impedance is equal to the resistance, and the power dissipated is calculated using the formula P = V2/R.
140 Medium
140 of 154
Analyzing +4 | -1
A series L−C−RL-C-R circuit is connected to an AC\mathrm{AC} voltage source. When LL is removed from the circuit, the phase difference between current and voltage is π4\frac{\pi}{4}. If instead CC is removed from the circuit, the phase difference is again π4\frac{\pi}{4} between current and voltage. The power factor of the circuit is
View answer and explanation
Correct option: B
1
Option B (1) is correct because when L and C are removed individually and the phase difference is π4\frac{\pi}{4}, it implies that the circuit is at resonance, making the power factor equal to 1.
141 Hard
141 of 154
Application +4 | -1
An L – C – R circuit of R=100 Ω is connected to an AC source 100 V, 50 Hz. The magnitude of phase difference between current and voltage is 30°. The power dissipated in the L – C – R circuit is
View answer and explanation
Correct option: B
86.6 W
Option B (86.6 W) is correct because the power dissipated is calculated using the formula P = Vrms * Irms * cos(phi), where cos(phi) is the power factor.
142 Hard
142 of 154
Application +4 | -1
An AC circuit consists of a 220Ω220 \Omega resistance and a 0.7H0.7 \mathrm{H} choke. The power absorbed from 220 V220 \mathrm{~V} and 50 Hz50 \mathrm{~Hz} source connected in this circuit, if the resistance and choke are joined in series is
View answer and explanation
Correct option: A
110 W110 \mathrm{~W}
Option A (110 W110 \mathrm{~W}) is correct because it is found by calculating the impedance of the series circuit, then the current, and finally the power absorbed by the resistor.
143 Medium
143 of 154
Understanding +4 | -1
A mixer of 100   Ω100\,\,{\mkern 1mu} \Omega resistance is connected to an A.C. source of 200 v and 50 cycles/sec. The value of average potential difference across the mixer will be
View answer and explanation
Correct option: D
zero
Option D is correct. The average potential difference across any AC source over a complete cycle is always zero.
144 Easy
144 of 154
Understanding +4 | -1
A transformer of frequency 60 Hz60 \mathrm{~Hz} and 120 V120 \mathrm{~V} input has 8:18: 1 turn ratio. The frequency of the output is
View answer and explanation
Correct option: D
60 Hz60 \mathrm{~Hz}
Option D (60 Hz) is correct because the frequency of the output voltage in a transformer is the same as the frequency of the input voltage.
145 Easy
145 of 154
Application +4 | -1
The ratio of turns in primary and secondary coils of a transformer is 1 : 20. The ratio of currents in primary and secondary coils will be
View answer and explanation
Correct option: B
0.834027778
Answer A (0.055555556) is incorrect. The correct answer is the inverse of the turns ratio, which is 20. The provided answer is the inverse of 18.
146 Medium
146 of 154
Analyzing +4 | -1
Assertion: A bulb connected in series with a solenoid is connected to ac source. If a soft iron core is introduced in the solenoid, the bulb will glow brighter.

Reason: On introducing soft iron core in the solenoid, the inductance increases.
View answer and explanation
Correct option: D
If assertion is false but reason is true.
Option D (If assertion is false but reason is true) is correct because introducing a soft iron core decreases the bulb's brightness, not increases it.
147 Medium
147 of 154
Application +4 | -1
In a uniform magnetic field of induction B a wire in the form of a semicircle of radius rr rotates about the diameter of the circle with angular frequency ω\omega. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R\mathrm{R}, the mean power generated per period of rotation is :
View answer and explanation
Correct option: B
(Bπr2ω)28R\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}}
Option B ((Bπr2ω)28R\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}}) is correct because it represents the mean power generated per period of rotation, derived from Faraday's law and the power formula.
148 Easy
148 of 154
Application +4 | -1
A series LCR circuit with R=20′Ω,L=1.5HR=20^{\prime} \Omega, L=1.5 \mathrm{H} and C=35\mathrm{C}=35 μF\mu \mathrm{F} is connected to a variable-frequency 200 V200 \mathrm{~V} ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power in Kw\mathrm{Kw} transferred to the circuit in one complete cycle?
View answer and explanation
Correct option: A
2
Answer A (2) is correct because at resonance, the impedance is equal to the resistance, and the average power is calculated using P=V2/RP = V^2/R.
149 Medium
149 of 154
Analyzing +4 | -1
Assertion: The alternating current lags behind the e.m.f. by a phase angle of π/2\pi / 2, when ac flows through an inductor.

Reason: The inductive reactance increases as the frequency of ac source decreases.
View answer and explanation
Correct option: C
If assertion is true but reason is false.
Option C is correct because the assertion is true, but the reason is false. In an inductor, current lags voltage by π/2\pi/2. Inductive reactance increases with the frequency of the AC source.
150 Hard
150 of 154
Application +4 | -1
A pure resistive circuit element XX when connected to an AC supply of peak voltage 200 V200 \mathrm{~V} gives a peak current of 5 A5 \mathrm{~A} which is in phase with the voltage. A second circuit element Y\mathrm{Y}, when connected to the same AC supply also gives the same value of peak current but the current lags behind by 90∘90^{\circ}. If the series combination of X\mathrm{X} and Y\mathrm{Y} is connected to the same supply, the rms valus of current is
View answer and explanation
Correct option: A
52 A\frac{5}{\sqrt{2}} \mathrm{~A}
Option A (52 A\frac{5}{\sqrt{2}} \mathrm{~A}) is correct because it is calculated by finding the resistance and reactance, then the impedance, and finally the RMS current.
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