Q. 60 The 聽tuning 聽circuit 聽in 聽an... [FREE SOLUTION]

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Chapter 32: Q. 60 (page 926)

The tuning circuit in an FM radio receiver is a series RLC circuit with a $0.200 m H$ inductor.
a. The receiver is tuned to a station at $104.3 m H Z$ . What is the value of the capacitor in the tuning circuit?
b. FM radio stations are assigned frequencies every $0.2 m H Z$ , but two nearby stations cannot use adjacent frequencies. What is the maximum resistance the tuning circuit can have if the peak current at a frequency of $103.9 m H Z$ , the closest frequency that can be used by a nearby station, is to be no more than $0.10 %$ of the peak current at $104.3 m H Z$ ? The radio is still tuned to $104.3 M H z$ , and you can assume the two stations have equal strength.

Short Answer

Expert verified

(a). The value of capacitor is $11.64 p F$ .

(b). The maximum resistance is $1.31 脳 10^{5} 惟$

Step by step solution

Part (a) Step 1: Given information

We are given that a tuning circuit in an FM radio receiver is a series RLC circuit with a $0.200 渭 H$ inductor. and The receiver is tuned to a station at $104.3 M H Z$ . and We need to find that What is the value of the capacitor in the tuning circuit?

Part (a) Step 2: Explanation

In this question we are given that $L = 0.200 渭 H$ $f = 104.3 M H Z$

we know that $蝇 = \frac{1}{\sqrt{L 脳 C}}$ for circuit.

$蝇 = 2 脳蟺脳 f$

now put the value of $蝇$ in equation.

localid="1650288071735" $2 脳蟺脳 f = \frac{1}{\sqrt{L 脳 C}} C = \frac{1}{4 脳蟺^{2} 脳 f^{2} 脳 L} C = \frac{1}{4 脳 3.14 脳 3.14 脳 104 . 3^{2} 脳 10^{6} 脳 0.2} C = 11.6 p F$

so , the value of capacitance is $11.6 k F$

Part (b) Step 1: Given information

We are given that tuning circuit in an FM radio receiver is a series RLC circuit with a $0.200 渭 H$ inductor. and We need to find that What is the maximum resistance the tuning circuit can have if the peak current at a frequency of $103.9 M H Z$ , the closest frequency that can be used by a nearby station, is to be no more than $10 %$ of the peak current at localid="1650288142554" $104.3 M H Z$ ? The radio is still tuned to localid="1650288151188" $104.3 M H Z$ , and you can assume the two stations have equal strength.

Part (b) Step 2: Explanation

We are given that $f = 104.3 m H Z$

We have to use the formula

$I_{p} = V_{P} / R I_{p} = V_{P} / X_{C} = V_{P} / X_{L}$

Now we have to find R so

role="math" localid="1650738106307" $\frac{V_{p}}{R} = 0.001 x \frac{V_{p}}{X_{L}} 0.001 R 鈥� = X_{L} 0.001 R 鈥� = 2 蟺蹿尝 R 鈥� = 1.31 x 10^{5} ohm$

So Maximum resistance is $R 鈥� = 1.31 x 10^{5} ohm$ .

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Short Answer

Step by step solution

Part (a) Step 1: Given information

Part (a) Step 2: Explanation

Part (b) Step 1: Given information

Part (b) Step 2: Explanation

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