An FM radio tuning circuit has a coil with an inductance of 0.0015 mH. What is the value of the capacitance if the set is tuned to 98 MHz? A) 1.8 *10^-⁶ $\mu$F B) 0.88 *10^-⁶$\mu$F C) 0.98 *10^-⁶ $\mu$F D) 12 *10^-⁶ $\mu$F

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Quizplus · Accepted Answer

The resonant frequency of an LC circuit is given by:$$f = \frac{1}{2\pi\sqrt{LC}}$$where:* f is the resonant frequency in Hz* L is the inductance in henrys* C is the capacitance in faradsWe are given that the resonant frequency is 98 MHz, which is equal to 98 x 10^6 Hz. We are also given that the inductance is 0.0015 mH, which is equal to 1.5 x 10^-6 H. Substituting these values into the equation above, we get:$$98 	imes 10^6 Hz = \frac{1}{2\pi\sqrt{1.5 	imes 10^{-6} H 	imes C}}$$Solving for C, we get:$$C = \frac{1}{4\pi^2(98 	imes 10^6 Hz)^2(1.5 	imes 10^{-6} H)} = 1.8 	imes 10^{-12} F$$Therefore, the value of the capacitance is 1.8 nF.

An FM Radio Tuning Circuit Has a Coil with an Inductance