Physics · Alternating currents, peak and RMS value of alternating current/voltage

An alternating voltage given as is applied to a capacitor of . The current readi

An alternating voltage given as \( \boldsymbol{V}= \) \( \mathbf{1 0 0} \sqrt{\mathbf{2}} \sin 100 t \quad \) is applied to a capacitor of \( 1 \mu F \). The current reading of the ammeter will be equal to \( \mathrm{mA} \)

  • A. 10
  • B. 20
  • C. 40
  • D. 80

Step-by-step solution

Given V = 100√2 sin(100t), peak voltage V0 = 100√2 V, ω = 100 rad/s, C = 1 μF = 10^-6 F. Capacitive reactance Xc = 1/(ωC) = 1/(100×10^-6) = 10^4 Ω. Peak current I0 = V0/Xc = (100√2)/10^4 = √2/100 A. RMS current Irms = I0/√2 = (√2/100)/√2 = 0.01 A = 10 mA. Ammeter reads RMS value, so answer is 10 mA.
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