Question

A 108-V resistor, a 0.200-mF capacitor, and a 5.42-mH inductor are connected in series to a generator whose voltage is 26.0 V. The current in the circuit is 0.141 A. Because of the shape of the current–frequency graph, there are two possible values for the frequency that corresponds to this current. Obtain these two values

Answer: f1=7.5x10^3 Hz , f2 = 3.11 x 10^3

Answer 1

impedance.
Zr = R = 108
Zc = 1 / (ωC)
Zl = ωL

Z = √[R² + (Zl - Zc)²]

V = iZ
V² = i²Z²
V² = i² [R² + (Zl - Zc)²]
26² = 0.141² ( 108² + [ω5.42*10⁻³ - 1 / (ω0.2*10⁻⁶)²])
34002 = 108² + [ω5.42*10⁻³ - 1 / (ω0.2*10⁻⁶)]²
22338.3 = [ω5.42*10⁻³ - 1 / (ω0.2*10⁻⁶)]²
149.46 = (ω5.42*10⁻³ - 1 / (0.2*10⁻⁶)
149.46ω = ω²5.42*10⁻³ - 1 / (0.2*10⁻⁶)
ω²5.42*10⁻³ - 149.46ω - 1 / (0.2*10⁻⁶) = 0
ω²5.42*10⁻³ - 149.46ω - 5*10⁶ = 0

Solve like a quadratic equation

ω = 4.71*10⁴
ω = -1.96*10⁴

ω = 2πf
f = ω / (2π)

f = 7500 Hz
f = -3110 Hz