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Alternating Current question

2022 · 25 Jul · Shift 1 · Q52
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  5. /2022 · 25 Jul · Shift 1 · Q52

Alternating Current question

2022 · 25 Jul · Shift 1 · Q52

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
To increase the resonant frequency in series LCR circuit,
  1. A
    source frequency should be increased.
  2. B
    another resistance should be added in series with the first resistance.
  3. C
    another capacitor should be added in series with the first capacitor.
  4. D
    the source frequency should be decreased.
View written solutionFree

Correct answer: C

  1. Resonant frequency of a series LCR circuit

For a series LCR circuit, the resonant frequency is

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}f0​=2πLC​1​

where:

  • LLL = inductance
  • CCC = capacitance

So, to increase f0f_0f0​, we must decrease the quantity LC\sqrt{LC}LC​, i.e. decrease either LLL or CCC.


  1. Check each option

Option A: source frequency should be increased

Changing the source frequency does not change the circuit’s natural resonant frequency. Resonant frequency depends only on LLL and CCC.

So, A is incorrect.

Option B: another resistance should be added in series with the first resistance

Resistance does not appear in

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}f0​=2πLC​1​

Hence adding resistance does not increase resonant frequency.

So, B is incorrect.

Option C: another capacitor should be added in series with the first capacitor

If capacitors are added in series, the equivalent capacitance decreases:

1Ceq=1C1+1C2\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}Ceq​1​=C1​1​+C2​1​

Thus,

Ceq<C1C_{\text{eq}} < C_1Ceq​<C1​

Since resonant frequency is

f0=12πLCeqf_0 = \frac{1}{2\pi\sqrt{L C_{\text{eq}}}}f0​=2πLCeq​​1​

a smaller capacitance gives a larger resonant frequency.

So, C is correct.

Option D: source frequency should be decreased

Again, source frequency does not determine the resonant frequency of the circuit.

So, D is incorrect.


  1. Final answer

The resonant frequency increases when the equivalent capacitance decreases, which happens by adding another capacitor in series.

C\boxed{\text{C}}C​

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