Answer:
instantaneous current will be 59.52 mA
Explanation:
Given data
oscillation F = 3.80 kHz = 3.80 ×[tex]10^{3}[/tex] Hz
voltage V = 3.80 V
inductance L = 3.50 mH = 3.50 ×[tex]10^{-3}[/tex] H
capacitor C = 290. nF = 290 ×[tex]10^{-9}[/tex] F
resistor R = 19.0 Ω
to find out
the instantaneous current
solution
we know that current I = V / Z
here Z = √(R²+(xl - xc)²)
so first we find xl = 2π×f×L = 2π×3800×3.50 ×[tex]10^{-3}[/tex]
xl = 83.52 ohm
and xc = 1 / 2π×f×C = 1 / 2π×3800× 290 ×[tex]10^{-9}[/tex]
xc = 144.497 ohm
so Z = √(R²+(xl - xc)²)
Z = √(19²+(83.52 - 144.497)²)
Z = 63.84
so that current will be V / Z
current = 3.80 / 63.84
current = 0.5952 A
so instantaneous current will be 59.52 mA
