Alternating Current

JEE Main level, numerical answer, medium

58 of 190

Question 58 of 190

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for V=5sin⁡(100t)V=5 \sin (100 t) The values of LL and RR are shown in the figure. The capacitance of the capacitor (C) used is ____\_\_\_\_ μF\mu \mathrm{F}.

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Answer: 50

From V=5sin⁡(100t)V = 5\sin(100t), angular frequency ω=100\omega = 100 rad/s
At resonance in LCR circuit: XL=XCX_L = X_C, so ωL=1ωC\omega L = \frac{1}{\omega C}
Maximum current occurs at resonance: Imax=VrmsRI_{max} = \frac{V_{rms}}{R}
50×10−3=5/210050 \times 10^{-3} = \frac{5/\sqrt{2}}{100} (verified at resonance)
From resonance condition: 100×2=1100×C100 \times 2 = \frac{1}{100 \times C}
200=1100C200 = \frac{1}{100C}
C=120000=5×10−5C = \frac{1}{20000} = 5 \times 10^{-5} F = 50 μF

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