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JEE • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 279 questions

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151 Medium
151 of 279
Understanding +4 | -1
The ratio of secondary to the primary turns in a transformer is 3:2\mathrm{3 : 2}. If the power output be PP, then the input power neglecting all losses must be equal to
View answer and explanation
Correct option: C
PP
Option C (PP) is correct because, in an ideal transformer (neglecting losses), the input power is equal to the output power.
152 Medium
152 of 279
Application +4 | -1
An ac ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the ac ammeter reads 3 A3 \mathrm{~A}. When another alternating current passes through the circuit, the ac ammeter reads 4 A4 \mathrm{~A}. Then find the reading of this ammeter (in A\mathrm{A}), if dc and ac flow through the circuit simultaneously.
View answer and explanation
Correct option: A
5
Option A (5) is correct because the AC ammeter reads the RMS value of the combined DC and AC currents, which is calculated to be 5A.
153 Medium
153 of 279
Application +4 | -1
A solenoid having inductance 4.0 H4.0 \mathrm{~H} and resistance 10 Ω10 \, \Omega is connected to a 4.0 V4.0 \mathrm{~V} battery at t=0\text{t}=0. Find

(a) the time constant
(b) the time elapsed before the current reaches 0.63 of its steady-state value
View answer and explanation
Correct option: A
(a) 0.40 s;0.40 \mathrm{~s}; (b) 0.40 s0.40 \mathrm{~s}
Option A ((a) 0.40 s;0.40 \mathrm{~s}; (b) 0.40 s0.40 \mathrm{~s}) is correct because the time constant and the time to reach 0.63 of steady-state current are both equal to L/R.
154 Medium
154 of 279
Understanding +4 | -1
Assertion (A) : Resonance frequency will decrease in L−C−RL-C-R series circuit if a dielectric slab is inserted in between the plates of the capacitor.

Reason (R) : By doing so, capacity of capacitor will increase.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A is correct because inserting a dielectric increases capacitance, which decreases the resonance frequency in an LCR circuit.
155 Medium
155 of 279
Remembering +4 | -1
The power factor of good choke coil is
View answer and explanation
Correct option: A
Nearly zero
Option A (Nearly zero) is correct because a good choke coil is designed to have very low resistance and high inductance, resulting in a power factor close to zero.
156 Medium
156 of 279
Analyzing +4 | -1
The value of current in two series LCRL C R circuits at resonance is same. Then
View answer and explanation
Correct option: C
For both the circuits XL/XCX_{{L}} / X_{{C}} must be same at that frequency
Option C is correct. At resonance, the inductive reactance (XLX_L) and capacitive reactance (XCX_C) are equal, so their ratio must be 1 for both circuits.
157 Medium
157 of 279
Application +4 | -1
In the circuit with R=6  ΩR = 6\,\,\Omega and impedance 10   Ω10\,\,\,\Omega. The power factor of circuit is
View answer and explanation
Correct option: B
0.6
Option B (0.6) is correct because the power factor is the ratio of resistance to impedance in an AC circuit.
158 Medium
158 of 279
Application +4 | -1
Find the effective value of current i=2+4cos⁡i=2+4 \cos 100πt100 \pi \mathrm{t}.
View answer and explanation
Correct option: A
232 \sqrt{3}
Option A (232 \sqrt{3}) is correct because it is the RMS value calculated by considering both the DC and AC components of the current.
159 Medium
159 of 279
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
The correct answer is C because electrical resonance requires both an inductor and a capacitor, and a pure capacitor or inductor has a power factor of zero, thus consuming no power.
160 Medium
160 of 279
Analyzing +4 | -1
Assertion (A) : In an AC\mathrm{AC} circuit, potential difference across the capacitor may be greater than the applied voltage.

Reason (R) : VC=IXCV_C=I X_C, whereas V=IZV=I Z and XCX_C can be greater than ZZ also.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A is correct because the potential difference across a capacitor in an AC circuit can be greater than the applied voltage due to resonance, and the reason correctly explains this with the impedance relationship.
161 Medium
161 of 279
Analyzing +4 | -1
A capacitor, an inductor and an electric bulb are connected in series to an AC supply of variable frequency. As the frequency of supply increased gradually, then the electric bulb is found to be
View answer and explanation
Correct option: C
Increase in brightness, reach a maximum and then decrease in brightness
Option C (Increase in brightness, reach a maximum and then decrease in brightness) is correct because the brightness is related to the power, which is maximum at resonance.
162 Medium
162 of 279
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. When the power factor is 1, XL = XC. Using the given power factors, the ratio of resistances can be found.
163 Medium
163 of 279
Understanding +4 | -1
Assertion: A sinusoidal alternating current does not show any average magnetic effects.

Reason: The average value of an alternating current is zero.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because a sinusoidal alternating current has zero average value, resulting in no average magnetic effect.
164 Medium
164 of 279
Application +4 | -1
A coil having an inductance of 1π\frac{1}{\pi} henry is connected in series with 300Ω300 \Omega.If 20 V20 \mathrm{~V} from a 200 cycle source is impressed across the combination, then the phase angle between, V and I\mathrm{I} is-
View answer and explanation
Correct option: D
tan⁡−1(43)\tan ^{-1}\left(\frac{4}{3}\right)
Option D (tan⁡−1(43)\tan ^{-1}\left(\frac{4}{3}\right)) is correct because the phase angle is calculated using the inductive reactance and resistance.
165 Medium
165 of 279
Understanding +4 | -1
The instantaneous value of current and emf in an A.C. circuit are I=12sin⁡314tI = \frac{1}{{\sqrt 2 }}\sin 314t amp and E=2sin⁡(314t−π6)V,E = \sqrt 2 \sin \left( {314t - \frac{\pi }{6}} \right)V, with respect to I) will be
View answer and explanation
Correct option: A
−π6rad- \frac{\pi }{6}\mathrm{ rad}
Option A (−π6rad- \frac{\pi }{6}\mathrm{ rad}) is correct because the phase difference is directly observed from the given equations.
166 Medium
166 of 279
Understanding +4 | -1
There is a current of 20 mA20 \mathrm{~mA} passing through an ideal inductor when connected to a 100 V,20 Hz100 \mathrm{~V}, 20 \mathrm{~Hz} supply. The power consumed is
View answer and explanation
Correct option: D
Zero
Option D is correct because an ideal inductor does not dissipate power; it only stores energy in its magnetic field.
167 Easy
167 of 279
Understanding +4 | -1
Statement I: The power factor of an inductor is zero.

Statement II: In the inductor, the emf and current differ in phase by π2\frac{\pi}{2}.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is
the correct explanation for Statement I
Option A (Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I) is correct because the power factor is the cosine of the phase angle between voltage and current, and in an inductor, the phase difference is π2\frac{\pi}{2}, making the power factor zero.
168 Medium
168 of 279
Application +4 | -1
An alternating current of rms value 10 A passed through 12 ohm resistor. The maximum potential difference across the resistor is [in Volts]
View answer and explanation
Correct option: C
169.98
Option C (169.98) is correct because the maximum potential difference is calculated by multiplying the maximum current by the resistance, where the maximum current is 2\sqrt{2} times the RMS current.
169 Medium
169 of 279
Remembering +4 | -1
Transformer is based upon the principle of
View answer and explanation
Correct option: B
mutual induction
Option B is correct because a transformer operates based on the principle of mutual induction between two coils.
170 Medium
170 of 279
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 is correct because when the dielectric is filled, the capacitance increases, impedance decreases, and thus the current through the resistor increases.
171 Medium
171 of 279
Understanding +4 | -1
The peak value of house hold line voltage is
View answer and explanation
Correct option: B
311 V311 \mathrm{~V}
Option B (311 V311 \mathrm{~V}) is correct because the peak voltage is calculated using the formula Vm=2Vrms V_m = \sqrt{2} V_{\text {rms }} where Vrms V_{\text {rms }} is 220 V220 \mathrm{~V}.
172 Medium
172 of 279
Application +4 | -1
Find the output voltage of an ideal transformer connected to a 220 V220 \mathrm{~V} A.C and number of turns in primary coil and secondary coil of transformer are 100 and 10 respectively.
View answer and explanation
Correct option: A
22 V
Option A (22 V) is correct because it is calculated using the transformer equation relating the voltage ratio to the turns ratio.
173 Medium
173 of 279
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 resistance R is found to be zero, leading to a power factor of zero in the circuit.
174 Medium
174 of 279
Understanding +4 | -1
When capacitor is connected to the circuit, the current, when key is pressed, is
View answer and explanation
Correct option: B
maximum
Option B (maximum) is correct because initially, the capacitor acts as a short circuit, allowing maximum current to flow.
175 Medium
175 of 279
Understanding +4 | -1
Statement I : In a series L−CL-C - RR circuit, the resonance can take place.

Statement II : Resonance takes place, if the inductive and capacitive reactances are equal and opposite.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
Option A is correct because resonance in a series L-C-R circuit occurs when inductive and capacitive reactances are equal and opposite, leading to minimum impedance.
176 Medium
176 of 279
Application +4 | -1
A 100μF100 \mu \mathrm{F} capacitor in series with a 40 Ω40 ~\Omega resistance is connected to a 110 V,60 Hz110 \mathrm{~V}, 60 \mathrm{~Hz} supply. What is the maximum current in the circuit?
View answer and explanation
Correct option: A
3.24 A3.24 \mathrm{~A}
Option A (3.24 A3.24 \mathrm{~A}) is correct because it is calculated by finding the peak voltage, impedance, and then the maximum current in the R-C circuit.
177 Medium
177 of 279
Application +4 | -1
In a series LRL R circuit XL=RX_L=R and power factor of the circuit is P1P_1. When capacitor with capacitance CC such that XL=XCX_L=X_C is put in series, the power factor becomes P2P_2. The ratio P1P2\frac{P_1}{P_2} is
View answer and explanation
Correct option: B
12\frac{1}{\sqrt{2}}
Option B (12\frac{1}{\sqrt{2}}) is correct because the power factor changes from 12\frac{1}{\sqrt{2}} to 1 when resonance occurs.
178 Medium
178 of 279
Application +4 | -1
An ideal choke takes a current of 10 ampere when connected to an A.C. supply of 125 volt and 50 Hz50 \mathrm{~Hz}. A pure resistor under the same conditions takes a current of 12.5 ampere. If the two are connected to an A.C. supply of 1002100 \sqrt{2} volt and 40 hertz, then find the current in a series combination of the above resistor and inductor.
View answer and explanation
Correct option: A
0.416666667
The correct answer is 10:00 AM because the current in the series combination of the resistor and inductor is calculated to be 10 A using the given parameters.
179 Medium
179 of 279
Understanding +4 | -1
QQ-factor can be increased by having a coil of:
View answer and explanation
Correct option: A
Large inductance, small ohmic resistance
Option A (Large inductance, small ohmic resistance) is correct because the Q-factor of a coil is given by Q=ωLRQ = \frac{{\omega L}}{R}, so a large inductance and small resistance will increase the Q-factor.
180 Medium
180 of 279
Understanding +4 | -1
A choke coil has negligible resistance. The alternating potential drop across it is 220 volt and the current is 5mA. The power consumed is.
View answer and explanation
Correct option: C
zero
Option C (zero) is correct because the average power consumed by a purely inductive choke coil in an AC circuit is zero.
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