NEET Previous Year Questions (2014-2024): Alternating Current | Physics Class 12 PDF Download

2024

Q1: In an ideal transformer, the turns ratio is N_P / N_S = 1/2. The ratio V_S : V_P is equal to (the symbols carry their usual meaning) :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 1 : 4        [2024]
Ans: (b)
An ideal transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils. The primary coil (with N_P turns) is connected to the input voltage (V_P), and the secondary coil (with N_S turns) delivers the output voltage (V_S).
The relationship between the input (primary) voltage and output (secondary) voltage in a transformer is determined by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, expressed by the formula:

In this case, the given turns ratio is N_P / N_S = 1 / 2. To find the correct expression for V_S / V_P, we need to invert the given ratio to reflect the relationship with respect to the voltages. This means:

Therefore, substituting into the voltage ratio equation:

This indicates that the secondary voltage (V_S) is twice the primary voltage (V_P). If we were to express the ratio V_S : V_P based on this, it would be: Option B 2 : 1
This is the correct answer as it represents that, in accordance with the transformer's turns ratio, the voltage in the secondary coil is twice that in the primary coil.

Q2: A 10μF capacitor is connected to a  210V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π=3.14) :

(a) 0.58 A
(b) 0.93 A
(c) 1.20 A
(d) 0.35 A                [2024]
Ans: (b)
Capacitive Reactance

2023

Q1: An ac source is connected in the given circuit. The value of $�$ϕ will be : 

(a) 60o
(b) 90o
(c) 30o
(d) 45o
Ans: (d)

Q2:  If Z₁ and Z₂ are the impedances of the given circuits (a) and (b) as shown in figures, then choose the correct option

(a) Z₁ < Z₂
(b) Z₁ + Z₂ = 20 Ω
(c) Z₁ = Z₂
(d) Z₁ >Z₂
Ans: (a)

Q3: The maximum power is dissipated for an ac in a/an:
(a) resistive circuit
(b) LC circuit
(c) inductive circuit
(d) capacitive circuit
Ans: (a)
Power dissipated is maximum of purely resistive circuit.

Q4: For very high frequencies, the effective impedance of the circuit (shown in the figure) will be:-

(a) 4 Ω
(b) 6 Ω
(c) 1 Ω
(d) 3 Ω
Ans: (d)
As frequency is very high
X_C ≈ 0
X_L → α
Effective circuit will be

Effective impedance of circuit will be = 3Ω

 Q5: The magnetic energy stored in an inductor of inductance  4 μH carrying a current of  2 A is :
(a) 4 mJ
(b) 8 mJ
(c) 8 μJ
(d) 4 μJ
Ans: (c)
The formula for calculating the magnetic energy stored in an inductor is given by:

where:
E is the energy stored (in joules, J),
L is the inductance of the inductor (in henrys, H),
I is the current flowing through the inductor (in amperes, A).
Plugging the given values into the formula:

Hence, the energy stored in the inductor is 8 μJ.

Q6: In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 μF, and resistance R is 100 Ω. The frequency at which resonance occurs is              
(a) 15.9 rad/s
(b) 15.9 kHz
(c) 1.59 rad/s
(d) 1.59 kHz
Ans: (d)
Solution: 


Q7: A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to AC mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?            
(a)  0.27 A
(b) 2.7 A
(c) 3.7 A
(d) 0.37 A
Ans: (a)
Solution:

For ideal transformer
P_input = P_output 
(VI)_in = 60
220 × I = 60
I = 0.27 A

Q8: An AC source is connected to a capacitor C. Due to a decrease in its operating frequency          
(a) Capacitive reactance decreases
(b) Displacement current increases
(c) Displacement current decreases
(d) Capacitive reactance remains constant
Ans: (c)
Solution: 

Since ω decreasing X_C will increase hence current will decrease. Also conduction current = displacement current.
Therefore displacement current will decrease.

Q9: The net impedance of the circuit (as shown in the figure) will be            
(a) 10√2 Ω
(b) 15 Ω
(c) 5√5
(d) 25 Ω
Ans: (c)
Solution:

2022

Q1: Given below are two statements
Statement I : In an a.c circuit, the current through a capacitor leads the voltage across it.
Statement II : In a.c circuit containing pure capacitance only, the phase difference between the current and voltage is π.
In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Statement I is incorrect but Statement II is correct
(b) Both Statement I and Statement II are correct
(c) Both Statement I and Statement II are incorrect
(d) Statement I is correct but Statement II is incorrect
Ans: (d)

Thus statement I is correct while II is incorrect.

Q2: An inductor of inductance 2 mH is connected to a 220 V, 50 Hz ac source. Let the inductive reactance in the circuit is X₁. If a 220 V dc source replace the ac source in the circuit, then the inductive reactance in the circuit is X₂, X₁ and X₂ respectively are :
(a) 0.628 Ω, infinity
(b) 6.28 Ω, zero
(c) 6.28 Ω, infinity
(d) 0.628 Ω, zero
Ans: (d)
We know, for A.C. Source

For D.C. Source
The inductor behaves as a closed circuit offering no resistance at all (at steady state) as w = 0 (For D.C.)

Q3: A standard filament lamp consumes 100 W when connected to 200 V ac mains supply. The peak current through the bulb will be :
(a) 2 A
(b) 0.707 A
(c) 1 A
(d) 1.414 A
Ans: (b)

Q4: A series LCR circuit with inductance 10 H, capacitance 10 µF, and resistance 50 Ω is connected to an AC source of voltage, V = 200sin(100t) volt. If the resonant frequency of the LCR circuit is  ν_o and the frequency of the AC source is ν, then             
(a)
(b)  
(c)  
(d)  

Ans: (a)
Solution:
 

Q5: The peak voltage of the AC source is equal to        
(a) The rms value of the AC source 
(b) √2 times the rms value of the AC source 
(c) 1/√2 times the rms value of the AC source 
(d) The value of the voltage supplied to the circuit 
Ans: (b)
Solution:
 

2021

Q1: A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by
V = V₀sinωt
The displacement current between the plates of the capacitor, would then be given by :
(a)

(b)

(c)

(d)

Ans: (b)
Given, V = V₀sinωt 
We know, q = CV
Now displacement current I_d is given by,

Q2: An inductor of inductance L, a capacitor of capacitance C, and a resistor of resistance 'R' are connected in series to an AC source of potential difference 'V' volts as shown in Fig. Potential difference across L, C, and R is 40 V, 10 V, and 40V, respectively. The amplitude of the current flowing through the LCR series circuit is 10√2A. The impedance of the circuit is:     
(a) 4Ω

(b) 5Ω
(c) 4√2Ω
(d) 5√2Ω
Ans: (b)
Solution:
I₀ = 10√2 A

Q3: A step-down transformer connected to an AC mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?    
(a) 2 A

(b) 4 A

(c) 0.2 A

(d) 0.4 A
Ans: (c)
Solution:

Q4: A series LCR circuit containing a 5.0 H inductor, 80 μF capacitor, and 40 Ω resistor is connected to a 230 V variable frequency AC source. The angular frequencies of the source at which power transferred to the circuit are half the power at the resonant angular frequency are likely to be    
(a) 46 rad/s and 54 rad/s

(b) 42 rad/s and 58 rad/s
(c) 25 rad/s and 75 rad/s
(d) 50 rad/s and 25 rad/s
Ans: (a)
Solution: 

2020

Q1: A 40 μF capacitor is connected to a 200 V. 50 Hz ac supply. The rms value of the current on the circuit is, nearly :
(a) 2.05 A
(b) 2.5 A
(c) 25.1 A
(d) 1.7 A
Ans: (b)

Q2: A series LCR circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is π/3. If instead C is removed from the circuit, the phase difference is again π/3 between current and voltage. The power factor of the circuit is :    
(a) 1.0
(b) –1.0
(c) Zero
(d) 0.5
Ans: (a)
Solution:
When the inductor alone is removed,
When the capacitor alone is removed,

Thus, for the original circuit,
Power factor = cos Δϕ = R/Z = R/R = 1

2019

Q1: Which of the following acts as a circuit protection device?    
(a) Conductor
(b) Inductor
(c) Switch
(d) Fuse
Ans: (d)
Solution:
Fuse wire has less melting point so when excess current flows, due to heat produced in it, it melts.

Q2: A parallel plate capacitor of capacitance 20 mF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively.    
(a) Zero, 60 μA
(b) 60 μA, 60 μA
(c) 60 μA, zero
(d) Zero, zero
Ans: (b)
Solution:
The capacitance of capacitor C = 20 mF
= 20 × 10^-6 F
Rate of change of potential i_c = 20 × 10^–6 × 3
= 60 × 10^–6 A
= 60 μA
As we know i_d = i_c = 60 mA

2018

Q1: The magnetic potential energy stored in a certain inductor is 25 mJ when the current in the inductor is 60 mA. This inductor is of inductance:    
(a) 0.138 H
(b) 138.88 H
(c) 1.389 H
(d) 13.89 H
Ans: (d)
Solution:
Energy stored in the inductor
Energy stored in the inductor

Q2:  In the circuit shown in the figure, the input voltage Vi is 20 V, V_BE = 0, and V_CE = 0. The values of I_B, I_C, and β are given by:

(a) I_B = 40 µA, IC = 10 mA, β = 250
(b) I_B = 25 µA, IC = 5 mA, β = 200
(c) I_B = 20 µA, IC = 5 mA, β = 250
(d) I_B = 40 µA, IC = 5 mA, β = 125
Ans: (d)
Solution:

Q3: An inductor 20 mH, a capacitor 100 μF, and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is    
(a) 0.79 W
(b) 0.43 W
(c) 2.74 W
(d) 1.13 W
Ans: (a)
Solution:

Q4: Figure shows a circuit that contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 m Heach, and an ideal battery with emf ε = 18 V. The current 'i' through the battery just after the switch closed is: