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In an AC circuit, an alternating voltage ε=2002sin⁡(100t)\varepsilon = 200\sqrt{2}\sin(100t) V is connected to a capacitor of capacity 1μF1 \mu\text{F}. The RMS value of the current in the circuit is:

A

100 mA

B

200 mA

C

20 mA

D

10 mA

Step-by-Step Solution

  1. Identify Parameters: Compare the given voltage equation ε=2002sin⁡(100t)\varepsilon = 200\sqrt{2}\sin(100t) with the standard form V=Vmsin⁡(ωt)V = V_m \sin(\omega t). Peak Voltage Vm=2002V_m = 200\sqrt{2} V. Angular Frequency ω=100\omega = 100 rad/s.
  • Capacitance C=1μF=1×10−6C = 1 \mu\text{F} = 1 \times 10^{-6} F.
  1. Calculate Reactance: The capacitive reactance is XC=1ωC=1100×10−6=110−4=104 ΩX_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \, \Omega .
  2. Calculate RMS Voltage: Vrms=Vm2=20022=200V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 V .
  3. Calculate RMS Current: Irms=VrmsXC=200104=0.02I_{rms} = \frac{V_{rms}}{X_C} = \frac{200}{10^4} = 0.02 A.
  4. Convert Units: 0.02 A=200.02 \text{ A} = 20 mA.
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