THE PROBLEM You have built an inductor by winding a wire around a cardboard tube
ID: 1585727 • Letter: T
Question
THE PROBLEM You have built an inductor by winding a wire around a cardboard tube. Its inductance is L-448 H. You want to put it into an LC circuit, using a variable capacitor to reach the full range of AM radio, which is 550 kHz to 1600 kHz. Calculate the required range for your variable capacitor. PAPER SOLUTION Solve the problem on paper first, including all four IDEA steps. You will become a better physicist that way! Have you finished your paper solution already? Oyes Ono INTERPRET Identify the true statement. A. The given AM radio range is for angular frequency. 7 B. The given AM radio range is for resonant frequency. C. None of the above. DEVELOP Derive an algebraic expression for the capacitance when the frequency is fExplanation / Answer
Given
L = 448*10^-6 H
the range of the frequency is f1 = 550 kHz to f2 = 1600 kHz
capacitance of the circuit range is say C1,C2 the lower and higher values
we know that the resonance frequency in rad/s is
W = 1/sqrt(L*C)
and we know W = 2pi*f
f = W/2pi
f = 1/(2pi*sqrt(L*C))
for f1 = 550 k Hz , C1 = 1/(2pi*sqrt(L))f1
The algebraic expression for capacitance with f is
C = 1/(2pi*sqrt(L))f
C1 = 1/(2pi*550*10^3* sqrt(448*10^-6)) F
C1 = 1.3671571407396*10^-5 F
C1 = 13.67157*10^-6 F
for f2 = 1600 k Hz , C2 = 1/(2pi*sqrt(L))f2
C2 = 1/(2pi*1600*10^3* sqrt(448*10^-6)) F
C2 = 4.6996026712923*10^-6 F
C2 = 4.69960*10^-6 F
so the range of the capacitance is C 1 = 13.67157*10^-6 F and C2 = 4.69960*10^-6 F
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