Respuesta :
Answer:
Explanation:
Given the following information,
Resistor of resistance R = 13.6Ω
Capacitor of capacitance C = 11.9-μF
C = 11.9 × 10^ -6 F
Inductor of inductance L = 19.1-mH
L = 19.1 ×10^-3 H
All this connected in series to a generator that generates Vrms= 117V
Vo = Vrms√2 = 117√2
Vo = 165.463V
a. Frequency for maximum current?
Maximum current occurs at resonance
I.e Xc = XL
At maximum current, the frequency is given as
f = 1/(2π√LC)
Then,
f = 1/(2π√(19.1×10^-3 × 11.9×10^-6)
f = 1/(2π√(2.2729×10^-7))
f = 1/(2π × 4.77 ×10^-4)
f = 333.83Hz
Then, the frequency is 333.83Hz.
b. Since we know the frequency,
Then, we need to find the capacitive and inductive reactance
Capacitive reactance
Xc = 1/2πfC
Xc = 1/(2π × 338.83 × 11.9×10^-6)
Xc = 1/ 0.024961
Xc = 40.1Ω
Also, Inductive reactance
XL = 2πfL
XL = 2π × 333.83 × 19.1×10^-3
XL = 40.1Ω
As expected, Xc=XL, resonance
Then, the impedance in AC circuit is given as
Z = √ (R² + (Xc—XL)²)
Z = √ 13.6² + (40.1-40.1)²)
Z = √13.6²
Z = 13.6 ohms
Then, using ohms las
V = IZ
Then, I = Vo/Z
Io = 165.46/13.6
Io = 12.17Amps
The current is 12.17 A
Answer:
a) Current is maximum at frequency, f₀ = 333.83 Hz
b) Maximum current = 12.17 A
Explanation:
Inductance, L = 19.1 mH = 19.1 * 10⁻³ H
Capacitance, C = 11.9 μF =11.9 * 10⁻⁶ F
a) Current is maximum at resonant frequency, f₀
[tex]f_{0} = \frac{1}{2\pi\sqrt{LC} }[/tex]
[tex]f_{0} = \frac{1}{2\pi\sqrt{11.9 * 10^{-6}* 19.1 * 10^{-3} } }[/tex]
[tex]f_{0} = 333.83 Hz[/tex]
b) Maximum value of the RMS current
[tex]V_{RMS} = 117 V\\V_{max} = \sqrt{2} V_{RMS}\\V_{max} = \sqrt{2} * 117\\V_{max} = 165.46 V[/tex]
[tex]I_{max} = \frac{V_{max} }{R} \\I_{max} = \frac{165.46}{13.6} \\I_{max} = 12.17 A[/tex]
