Problem 21.70 Constants Part A The variable capacitor in the tuner of an AM radi

Question

Problem 21.70 Constants Part A The variable capacitor in the tuner of an AM radio has a capacitance of 2500 pF when the radio is tuned to a station at 560 kHz What must be the capacitance for a station at 1300 kHz? Express your answer to two significant figures and include the appropriate units. ? c= | Value Units SubmitP Previous Answers Request Answer incorrect; Try Again; 3 attempts remaining Part B What is the inductance (assumed constant)? Express your answer using two significant figures. Submit Previous Answers Request Answer x Incorrect; Try Again; 3 attempts remaining

Explanation / Answer

Solution

Let us go to the basics first.

The formula for the resonant frequency of an LC circuit is

1/f = 2*pi*sqrt(L*C)

where: f = frequency [Hz], L = inductance [H], C = capacitance [F]

So the ratio of two frequencies f1, f2 in terms of two capacitors C1, C2 (assuming a constant inductance)
f1/f2 = sqrt(C2/C1)
C2 = C1*(f1/f2)2
Thus, C2= 2500pF*(560/1300)2
C2 = 463.9 pF = 460 pF (2 significant figures) (Answer a)

The inductance is found using the original equation, rearranged
L = 1/( (2*pi*f)2*C ) = 1/( (2*pi*560e3)2*2500e-12)
Thus, L = 32.3 uH = 32 uH (2 significant figures) (Answer b)

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