Answer:
2.36 x 10^6 Hz
Explanation:
We know ;,
Iamp = 1.8mA
L = 0.45mH
VL = 12V
VL = IwL
VL = I x 2π x f x L
f = VL / (I x 2π x L)
f = 12 V/ (1.8mA x 2π x 0.45mH)
f = 12/(1.8x10^-3 x 2π x 0.45 x 10^-3)
f = 2.36 x 10^6 Hz
The frequency that is required in this inductor circuit is equal to 2.4 MHz.
Given the following data:
To determine the frequency that is required:
According to Ohm's law, the voltage in an alternating current (AC) circuit is given by the formula:
[tex]V=IX_L = I\omega L[/tex]
Where:
Making [tex]\omega[/tex] the subject of formula, we have:
[tex]\omega = \frac{V}{IL}[/tex]
Substituting the given parameters into the formula, we have;
[tex]\omega = \frac{12}{0.450 \times 10^{-3} \times 1.8 \times 10^{-3}}\\\\\omega = \frac{12}{8.1 \times 10^{-7}} \\\\\omega = 1.48 \times 10^{7}\;rad/s[/tex]
Now, we can determine the frequency that is required:
[tex]Frequency = \frac{\omega}{2\pi} \\\\Frequency = \frac{1.48 \times 10^{7}}{6.284} \\\\Frequency = 2.4 \times 10^{6}\;Hertz[/tex]
Note: 1 MHz = [tex]1 \times 10^{6}[/tex] Hertz.
Frequency = 2.4 MHz.
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