An ac voltage V=220 sin (2×10³t) volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given: L=10 mH, C=25 μF, R=100 Ω)

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(2) 2.2 A. Read ω=2×10³ rad s⁻¹ from the source. X_L=ω L=2×10³×10×10⁻³=20 Ω. X_C=1/(ω C)=1/(2×10³×25×10⁻⁶)=1/(0.05)=20 Ω. The two reactances are equal, so the circuit sits exactly at resonance and Z=R=100 Ω. The amplitude of the current is i₀=(V₀)/Z=(220)/(100)=2.2 A.

An ac voltage V=220 sin (2×10³t) volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given: L=10 mH, C=25 μF, R=100 Ω)

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