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°?

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

i think the answer is b

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it