Question

A series RLC circuit with a resistance of 121.0 Ω has a resonance angular frequency of 5.1 ✕ 105 rad/s. At resonance, the voltages across the resistor and inductor are 60.0 V and 40.0 V, respectively.
At what frequency does the current in the circuit lag the voltage by 45°?

Answer 1

Answer:

Explanation:

At resonance ω₀L = 1 / ω₀C , L is inductance and C is capacitance .

C = 1 /  ω₀²L , ω₀ = 5.1 x 10⁵ . ( given )

voltage over resistance = R I , R is resistance and I is current

voltage over inductance   =  Iω₀L

R I /  Iω₀L = 60 / 40

R / ω₀L = 3 / 2

L = 2 R / 3 ω₀

= 2 x 121 / 3 x 5.1 x 10⁵

= 15.81 x 10⁻⁵

C = 1 /  ω₀²L

= 1 / (5.1 x 10⁵)² x 15.81 x 10⁻⁵

= .002432 x 10⁻⁵

= 24.32 x 10⁻⁹ F

Let the angular frequency required be ω

Tan 45 = (ωL - 1 / ωC) / R

ωL - 1 / ωC = R

ω²LC - 1 = R ωC

ω²LC = 1 + R ωC

ω² x 15.81 x 10⁻⁵ x 24.32 x 10⁻⁹ = 1 + 121 x ω x 24.32 x 10⁻⁹

ω² x 384.5 x 10⁻¹⁴ = 1 + 2942.72 x10⁻⁹ω

ω² - 7.65 x 10⁶ ω - 1 = 0

ω = 7.65 x 10⁶

frequency = 7.65 x 10⁶ / 2π

= 1.22 x 10⁶ Hz