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NEET • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Mixed difficulty

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91 Hard
91 of 154
Application +4 | -1
A resistance of 20 ohms is connected to a source of an alternating potential V=220sin⁡[100 π t]V = 220\sin \left[ {100\,\pi \,t} \right]. The time taken by the current to change from its peak value to r.m.s. value is [in sec]
View answer and explanation
Correct option: D
2.5×10−32.5 \times {10^{ - 3}}
Option D (2.5×10−32.5 \times {10^{ - 3}}) is correct because it involves finding the time difference between the peak and RMS values of a sinusoidal current.
92 Medium
92 of 154
Understanding +4 | -1
A bulb and a capacitor are in series with an ac source. On increasing frequency how will glow of the bulb change
View answer and explanation
Correct option: B
The glow increases
Option B is correct. Increasing the frequency decreases the capacitive reactance, allowing more current to flow and increasing the bulb's glow.
93 Hard
93 of 154
Application +4 | -1
An AC circuit having an inductor and a resistor in series draws a power of 560 W560 \mathrm{~W} from an AC source marked 210 V−60 Hz210 \mathrm{~V}-60 \mathrm{~Hz}. The power factor of the circuit is 0.8, the impedance of the circuit and the inductance of the inductor is
View answer and explanation
Correct option: C
63 Ω,0.1 H63 \, \Omega, 0.1 \, \mathrm{H}
Option C (63 Ω,0.1 H63 \, \Omega, 0.1 \, \mathrm{H}) is correct because it is calculated by finding the impedance from power and voltage, and then inductance from impedance and power factor.
94 Medium
94 of 154
Understanding +4 | -1
Same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the emf of AC is increased, the effect on the value of the current
View answer and explanation
Correct option: D
Decreases in the first circuit and increases in the other
Option D is correct because increasing the frequency decreases the current in an inductive circuit (due to increased inductive reactance) and increases the current in a capacitive circuit (due to decreased capacitive reactance).
95 Medium
95 of 154
Understanding +4 | -1
A resistor R, an inductor L and a capacitor C are connected in series to an oscillator of frequency nn, if the resonant frequency is nr{n_r} , then the current lags behind voltage, when
View answer and explanation
Correct option: D
n>nrn > {n_r}
Option D (n>nrn > {n_r}) is correct because when the frequency is greater than the resonant frequency, the inductive reactance dominates, causing the current to lag behind the voltage.
96 Medium
96 of 154
Analyzing +4 | -1
Assertion (A)If the frequency of the applied ACAC is doubled, then the power factor of a series R−LR-L circuit decreases.

Reason (R) Power factor of series R-L circuit s given' by

cos⁡θ=2RR2+ω2 L2\cos \theta=\frac{2 {R}}{\sqrt{{R}^2+\omega^2 {~L}^2}}
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. Increasing the frequency in an R-L circuit decreases the power factor. The correct formula for the power factor is cos⁡θ=RR2+(ωL)2\cos \theta = \frac{{R}}{{\sqrt{{R^2 + (\omega L)^2}}}}
97 Medium
97 of 154
Analyzing +4 | -1
A series R-C circuit is connected to AC Voltage source. Consider two cases: (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR{I_R} through the resistor and voltage Vc{V_c} across the capacitor are compared in the two cases. Which of the following is/are true?
View answer and explanation
Correct option: B
IRA<IRBI_R^A < I_R^B
Option B (IRA<IRBI_R^A < I_R^B) is correct because introducing a dielectric increases capacitance, leading to a lower impedance and thus a higher current.
98 Medium
98 of 154
Remembering +4 | -1
In step-up transformer, relation between number of turns in primary (Np)\left( {{N_p}} \right) and number of turns is secondary (Ns)\left( {{N_s}} \right) coils is
View answer and explanation
Correct option: A
Ns{N_s} is greater than Np{N_p}
Option A (Ns{N_s} is greater than Np{N_p}) is correct because in a step-up transformer, the number of turns in the secondary coil is greater than the number of turns in the primary coil to increase the voltage.
99 Medium
99 of 154
Understanding +4 | -1
Which of the following quantities remains constant in a step-down transformer ?
View answer and explanation
Correct option: C
Power
Option C (Power) is correct because, ideally, a transformer conserves power; the power in the primary coil equals the power in the secondary coil.
100 Easy
100 of 154
Understanding +4 | -1
In an ideal transformer, the turns ratio is NPNS=12.\frac{N_P}{N_S}=\frac{1}{2}. The ratio VS:VPV_S: V_P is equal to (the symbols carry their usual meaning):
View answer and explanation
Correct option: A
2:12: 1
Option A (2: 1) is correct because the voltage ratio is the inverse of the turns ratio, so VS/VP=NS/NP=2/1V_S/V_P = N_S/N_P = 2/1.
101 Medium
101 of 154
Application +4 | -1
In an L – C – R circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to
View answer and explanation
Correct option: C
L/2
Option C (L/2) is correct because the resonant frequency is inversely proportional to the square root of the product of inductance and capacitance.
102 Easy
102 of 154
Application +4 | -1
In an ideal transformer, the voltage is stepped down from 11 kV to 220 V. If the primary current be 100 A, the current in the secondary should be
View answer and explanation
Correct option: C
0.5 kA
Option A (5 kA) is correct because in an ideal transformer, the power is conserved, so VPIP=VSISV_P I_P = V_S I_S, which gives IS=(VPIP)/VS=(11000∗100)/220=5000A=5kAI_S = (V_P I_P) / V_S = (11000 * 100) / 220 = 5000 A = 5 kA.
103 Medium
103 of 154
Remembering +4 | -1
In a series LCRL C R circuit, at resonance, the bandwidth frequency can be represented as
View answer and explanation
Correct option: D
R2πL\frac{R}{2 \pi L}
Option D (R2πL\frac{R}{2 \pi L}) is correct because the bandwidth frequency in a series LCR circuit at resonance is given by the formula Δf=R2πL\Delta f = \frac{R}{2 \pi L}.
104 Medium
104 of 154
Understanding +4 | -1
The power is transmitted from a power house on high voltage ac because
View answer and explanation
Correct option: B
It is more economical due to less power wastage
Option B (It is more economical due to less power wastage) is correct because transmitting power at high voltage reduces current, thus reducing I2RI^2R losses in the transmission lines.
105 Medium
105 of 154
Analyzing +4 | -1
An ac source of angular frequency ω\omega is fed across a resistor rr and a capacitor CC in series. The current in the circuit measured is ll. If now the frequency of source is changed to ω/3\omega / 3 (but maintaining the same voltage), the current in the circuit is found to reduce by a factor of three. The power factor of the circuit at the original frequency ω\omega is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the problem states the current reduces by a factor of 3 when the frequency is reduced to ω/3\omega/3, implying resonance at the original frequency.
106 Medium
106 of 154
Remembering +4 | -1
The power factor varies between
View answer and explanation
Correct option: C
0 to 1
Option C (0 to 1) is correct because the power factor is the cosine of the phase angle between voltage and current, and cosine values range from 0 to 1.
107 Easy
107 of 154
Application +4 | -1
The current flowing in a step down transformer 220 V to 22 V having impedance 220 &Omega; is
View answer and explanation
Correct option: B
0.1A
Option B (0.1A) is correct because it is calculated using Ohm's law on the secondary side of the transformer.
108 Hard
108 of 154
Application +4 | -1
An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source 200 V, 50 Hz, a choke has to be used in series with it. The should have an inductance of
View answer and explanation
Correct option: D
1.1 H
Option D (1.1 H) is correct because the inductance is calculated based on the required impedance to limit the current through the bulb when used with a 200V source.
109 Medium
109 of 154
Application +4 | -1
An inductive circuit contains a resistance of 10 ohm and an inductance of 2.0 henry. If an ac voltage of 120 volt and frequency of 60 Hz is applied to this circuit, the current in the circuit would be nearly
View answer and explanation
Correct option: B
0.16 amp
Option A (0.32 amp) is correct because the current is calculated by dividing the voltage by the impedance of the RL circuit.
110 Hard
110 of 154
Application +4 | -1
In a series L−C−RL-C-R circuit, C=10−11 F,L=10−5HC=10^{-11} \mathrm{~F}, L=10^{-5} \mathrm{H} and R=100ΩR=100 \Omega, when a constant DC voltage EE is applied to the circuit, the capacitor acquires a charge 10−9C10^{-9} \mathrm{C}. The DC\mathrm{DC} source is replaced by a sinusoidal voltage source in which the peak voltage E0E_0 is equal to the constant DC\mathrm{DC} voltage EE. At resonance, the peak value of the charge acquired by the capacitor will be
View answer and explanation
Correct option: C
10−10C10^{-10} \mathrm{C}
Option D (10−8C10^{-8} \mathrm{C}) is correct because at resonance, the impedance is minimized, leading to a larger current and thus a larger charge on the capacitor.
111 Hard
111 of 154
Application +4 | -1
In an LCRL C R circuit, the resonating frequency is 200  kHz\,{kHz}. If the value of LL is doubled and value of CC is decreased to 18\frac{1}{8} times of its initial values, then the new resonating frequency in kHz{kHz} will be
View answer and explanation
Correct option: A
400
Option A (400 kHz) is correct because the resonating frequency is inversely proportional to the square root of the product of inductance and capacitance.
112 Hard
112 of 154
Application +4 | -1
The tuning circuit of a radio receiver has a resistance has a resistance of 50  Ω,50\,\,\Omega , an inductance of and 10 mH10\,mH a variable capacitance C.  A  1  MHzC.\,\,A\,\,1\,\,MHz radio wave produces a potential difference of 0.1 mV.0.1\,mV. The value of capacitance to produce the resonance will be
View answer and explanation
Correct option: A
2.5 pF2.5\,pF
Option A (2.5 pF2.5\,pF) is correct because it is the capacitance value that results in resonance at the given frequency and inductance.
113 Medium
113 of 154
Application +4 | -1
The sun delivers 103  w/m2{10^3}\,\,w/{m^2} of electromagnetic flux incident on a roof of dimensions 8  m  ×  20  m,8\,\,m\,\, \times \,\,20\,\,m, will be
View answer and explanation
Correct option: C
1.6×105  W1.6 \times {10^5}\,\,W
Option C (1.6×105 W1.6 \times {10^5}\,W) is correct because the total power is the intensity multiplied by the area.
114 Medium
114 of 154
Application +4 | -1
In an LCR circuit R=100 ohm. When capacitance C is removed, the current lags behind the voltage by π/3. When inductance L is removed, the current leads the voltage by π/3. The impedance of the circuit is
View answer and explanation
Correct option: B
100 ohm
Option B (100 ohm) is correct because at resonance, the impedance is equal to the resistance.
115 Easy
115 of 154
Remembering +4 | -1
The peak voltage of the ac source is equal to:
View answer and explanation
Correct option: D
2\sqrt 2 times the rms value of the ac source
Option D (2\sqrt 2 times the rms value of the ac source) is correct because the peak voltage is related to the RMS voltage by the factor of 2\sqrt{2}.
116 Medium
116 of 154
Application +4 | -1
A resistance of 40 ohm and an inductance of 95.5 millihenry are connected in series in a 50 cycles/second ac circuit. The impedance of this combination is very nearly
View answer and explanation
Correct option: C
50 ohm
Option C (50 ohm) is correct because the impedance is calculated using the formula Z=R2+XL2Z = \sqrt{R^2 + X_L^2}.
117 Medium
117 of 154
Understanding +4 | -1
Statement (A): If current varies sinusoidally the average power consumed in a cycle is zero.

Statement (B): If current varies sinusoidally the average current is zero in a complete cycle.
View answer and explanation
Correct option: C
(A) is false but (B) is true
Option C ((A) is false but (B) is true) is correct because the average power consumed in a cycle is not always zero if current varies sinusoidally, but the average current is zero.
118 Medium
118 of 154
Understanding +4 | -1
In an LRLR-circuit time constant is that time in which current grows from zero to the value (where I0{I_0} is the steady state current)
View answer and explanation
Correct option: A
0.63  I00.63\,\,{I_0}
Option A (0.63  I00.63 \,\,{I_0}) is correct because the current grows to approximately 63% of its steady-state value in one time constant in an LR circuit.
119 Medium
119 of 154
Understanding +4 | -1
Statement (A): The power consumed in an electric circuit is never negative
Statement (B): The average power consumed in an electric circuit is P=V2R=I2R{P}=\frac{{V}^2}{{R}}={I}^2{R}
View answer and explanation
Correct option: A
Both (A) and (B) are true
Option A (Both (A) and (B) are true) is correct because power consumed in an electric circuit is never negative, and the average power is given by P=V2R=I2R{P}=\frac{{V}^2}{{R}}={I}^2{R}.
120 Medium
120 of 154
Application +4 | -1
The instantaneous values of current and voltage in an ac circuit are i=100sin⁡ 314 t amp and e=200  in  (314  t+π/3)Ve = 200\;in\;\left( {314\;t + \pi /3} \right)V , respectively. If the resistance is 1Ω1\mathrm{\Omega } , then the reactance of the circuit will be
View answer and explanation
Correct option: B
3  Ω\sqrt 3 \;\mathrm{\Omega }
Option B (3  Ω\sqrt 3 \;\mathrm{\Omega }) is correct because the phase difference is used to calculate the reactance.
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