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MHT CET 2022 (11 Aug, Shift 2) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2022 (11 Aug, Shift 2)

Q47MHT CET 2022 · 11 AUG · SHIFT 2Easy

Pure R, L and C Circuits

A capacitor of capacitance 50 μF50\ \mu F is connected to a.c. source e=220sin⁡50te = 220 \sin 50t (ee in volt, tt in second). The value of peak current is

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Answer: (c)

  1. Peak current I0=e0XC=e0 ωCI_0 = \frac{e_0}{X_C} = e_0\,\omega C.
  2. e0=220e_0 = 220 V, ω=50\omega = 50 rad/s, C=50×10−6C = 50 \times 10^{-6} F.
  3. I0=220×50×50×10−6=0.55I_0 = 220 \times 50 \times 50 \times 10^{-6} = 0.55 A
  4. So the answer is (c) 0.550.55 A.
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