Question

A series RLC circuit is connected to a 3.80 kHz oscillator with a peak voltage of 3.80 V. It consists of a 3.50 mH inductor, a 290. nF capacitor, and a 19.0 ? resistor. If ? = ?0, what is the instantaneous current i?

Answer 1

instantaneous current will be 59.52 mA

Step-by-step explanation:

Given data

oscillation F = 3.80 kHz = 3.80 ×
`10^(3)` Hz

voltage V = 3.80 V

inductance L = 3.50 mH = 3.50 ×
`10^(-3)` H

capacitor C = 290. nF = 290 ×
`10^(-9)` 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 ×
`10^(-3)`

xl = 83.52 ohm

and xc = 1 / 2π×f×C = 1 / 2π×3800× 290 ×
`10^(-9)`

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