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PhysicsNEET 2024AC through R, L, C and LCR circuits

Alternating Current

A10μFA 10 \mu{}F capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (take π=3.14\pi = 3.14):

  1. A.0.93 A✓
  2. B.1.20 A
  3. C.0.35 A
  4. D.0.58 A

Solution

XC=12πfC=12×3.14×50×10−5≈318X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 10^{-5}} \approx 318 Ω\Omega. Peak current =V0XC=2102318≈0.93= \frac{V_0}{X_C} = \frac{210\sqrt2}{318} \approx 0.93 A.

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