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1.(JEE Main 2024 (Online) 6th April Morning Shift )

Given below are two statements :

Statement I : In an LCR series circuit, current is maximum at resonance.

Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source.

In the light of the above statements, choose the correct from the options given below :

A.Statement I is true but Statement II is false

B.Statement I is false but Statement II is true

C.Both Statement I and Statement II are true

D.Both Statement I and Statement II are false

Correct option is (c)

Statement I : True

Statement II : True

Current in purely resistive circuit is equal to current in LCR circuit with same resistance at resonance, otherwise more.

   

2.(JEE Main 2024 (Online) 5th April Evening Shift )

A series LCR circuit is subjected to an ac signal of 200   V , 50   Hz . If the voltage across the inductor ( L = 10   mH ) is 31.4   V , then the current in this circuit is _______.

A.10A

B.10mA

C.68A

D.63A

Correct option is (a)

V L = I ( ω L ) = 31.4 I = 31.4 2 × 3.14 × 50 × 10 × 10 3 = 10   A

   

3.(JEE Main 2024 (Online) 4th April Evening Shift )

Match List I with List II

LIST I LIST II
A. Purely capacitive circuit I. JEE Main 2024 (Online) 4th April Evening Shift Physics - Alternating Current Question 10 English 1
B. Purely inductive circuit II. JEE Main 2024 (Online) 4th April Evening Shift Physics - Alternating Current Question 10 English 2
C. LCR series at resonance III. JEE Main 2024 (Online) 4th April Evening Shift Physics - Alternating Current Question 10 English 3
D. LCR series circuit IV. JEE Main 2024 (Online) 4th April Evening Shift Physics - Alternating Current Question 10 English 4

Choose the correct answer from the options given below:

A.A-IV. B-I, C-III, D-II

B.A-I. B-IV, C-II, D-III

C.A-IV. B-I, C-II, D-III

D.A-I. B-IV, C-III, D-II

Correct option is (b)

For pure capacitive circuit, I lead by 90 to V

For pure inductive circuit, V lead by 90 to I

At series LCR resonance, I and V are in same phase.

For LCR series circuit, V and I may suffer some phase difference.

   

4.(JEE Main 2024 (Online) 1st February Morning Shift )

increased by 2   L

A.In series LCR circuit, the capacitance is changed from C to 4 C . To keep the resonance frequency unchanged, the new inductance should be:

B.reduced by 1 4   L

C.reduced by 3 4   L

D.increased to 4   L

Correct option is (c)

The resonance frequency f 0 of an LCR circuit is given by:

f 0 = 1 2 π L C

where:

  • L is the inductance.
  • C is the capacitance.

To keep the resonance frequency unchanged when the capacitance is changed from C to 4 C , we must adjust the inductance L to a new value L such that:

1 2 π L C = 1 2 π L 4 C

Now solving for L :

L C = 4 C L

Squaring both sides, we have:

L C = 4 C L

Dividing both sides by 4 C , we get:

L 4 = L

Thus, the new inductance L is one fourth of the original inductance L . Therefore, to achieve the same resonance frequency with the capacitance increased to 4 C , the inductance should be reduced to a quarter of its initial value:

L = L 4

This means we have reduced the inductance by 3 4 L , so the correct answer is:

Option C: reduced by 3 4 L

   

5.(JEE Main 2024 (Online) 27th January Evening Shift )

A series LCR circuit with L = 100 π mH , C = 10 3 π F and R = 10 Ω , is connected across an ac source of 220   V , 50   Hz supply. The power factor of the circuit would be ________.

Correct answer is 1

X c = 1 ω C = π 2 π × 50 × 10 3 = 10 Ω X L = ω L = 2 π × 50 × 100 π × 10 3 = 10 Ω X C = X L ,  Hence, circuit is in resonance   power factor  = R Z = R R = 1