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You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor.What frequency is required?

Respuesta :

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance [tex]L=0.450H=0.45\times 10^{-3}H[/tex]

Current in the inductor [tex]i=1.90mA=1.90\times 10^{-3}A[/tex]

Voltage v = 13 volt

Inductive reactance of the circuit [tex]X_l=\frac{v}{i}[/tex]

[tex]X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm[/tex]

We know that

[tex]X_l=\omega L=2\pi fL[/tex]

[tex]2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10[/tex]

f = 2421.127 kHz