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The AM radio band covers the frequency range from 520 kHz to 1610 kHz. Assuming a fixed inductance in a simple LC circuit, what ratio of capacitance is necessary to cover this frequency range? That is, what is the value of Ch/Cl, where Ch is the capacitance for the highest frequency and Cl is the capacitance for the lowest frequency? a. 3.37 b. 0.104 c. 0.568 d. 1.76 e. 12.3

User TLW
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The ratio of capacitance for the highest frequency (Ch) to the lowest frequency (Cl) in a fixed inductance LC circuit covering the AM radio band is approximately 12.3. Hence the correct option is e.

In a simple LC circuit used for tuning in the AM radio band, the frequency range spans from 520 kHz to 1610 kHz. The formula for resonant frequency in an LC circuit is given by f = 1 / (2π√(LC)), where f is the frequency, L is the inductance, and C is the capacitance. Since the inductance is fixed, the ratio of capacitance for the highest frequency (Ch) to the lowest frequency (Cl) can be expressed as Ch/Cl = √(f_h/f_l).

Substituting the given frequency values, we find that Ch/Cl is approximately 12.3. Therefore, the answer is 12.3, corresponding to option e. It indicates that the capacitance required for the highest frequency is about 12.3 times that of the capacitance needed for the lowest frequency in order to cover the entire AM radio band. Hence the correct option is e.

User Amarnath
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