For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF . If resistance of $200 \Omega$ and 100 mH inductor is being used in the given circuit. The frequency of ac source is $\_\_\_\_$ $\times 10^{3} \mathrm{~Hz}$. (given $\pi^{2}=10$ )

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Alternating Current

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF . If resistance of

200Ω

and 100 mH inductor is being used in the given circuit. The frequency of ac source is

____

\times 1 0^{3} Hz

. (given

π^{2} = 10

)

Answer: 10

At resonance the current is maximum, so

ω_{0} = \frac{1}{L C}

and

f_{0} = \frac{1}{2 π L C}

. With

L = 100 \times 1 0^{- 3}

H and

C = 2.5 \times 1 0^{- 9}

L C = 2.5 \times 1 0^{- 10}

. Using

π^{2} = 10

f_{0}^{2} = \frac{1}{4 π ^{2} L C} = \frac{1}{4 \times 10 \times 2.5 \times 1 0 ^{- 10}} = \frac{1}{1 0 ^{- 8}} = 1 0^{8} \Rightarrow f_{0} = 1 0^{4} Hz = 10 \times 1 0^{3} Hz

The stated resistance does not enter the resonant-frequency calculation.

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer