Problem 24.36 A 0.21 H inductor is connected to an ac generator with an rms volt

Question

Problem 24.36

A 0.21 H inductor is connected to an ac generator with an rms voltage of 12 V .

Part A

For what range of frequencies will the rms current in the circuit be less than 1.0 mA ?

Express your answer using two significant figures.

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Problem 24.52

Part A

What is the phase angle in an RLC circuit with R = 9.9 k , C = 1.4 F , and L = 240 mH ? The generator supplies an rms voltage of 100 V at a frequency of 60.0 Hz .

Express your answer using two significant figures.

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Problem 24.68

Part A

Find the frequency at which a 35 F capacitor has the same reactance as a 35 mH inductor.

Express your answer using two significant figures.

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Part B

What is the resonance frequency of an LC circuit made with this inductor and capacitor?

Express your answer using two significant figures.

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Problem 24.36

A 0.21 H inductor is connected to an ac generator with an rms voltage of 12 V .

Part A

For what range of frequencies will the rms current in the circuit be less than 1.0 mA ?

Express your answer using two significant figures.

f>   Hz  

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Incorrect; Try Again; 5 attempts remaining

Problem 24.52

Part A

What is the phase angle in an RLC circuit with R = 9.9 k , C = 1.4 F , and L = 240 mH ? The generator supplies an rms voltage of 100 V at a frequency of 60.0 Hz .

Express your answer using two significant figures.

=     

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Problem 24.68

Part A

Find the frequency at which a 35 F capacitor has the same reactance as a 35 mH inductor.

Express your answer using two significant figures.

f = Hz

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Part B

What is the resonance frequency of an LC circuit made with this inductor and capacitor?

Express your answer using two significant figures.

f =   Hz

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Explanation / Answer

(1) We Know that impedance of the inductor (XL) = 2Pi*f*L
where f is the frequency
XL = 2*Pi*fL = 2*Pi*f*(0.21*10-6)
We know that
I = V/Z = 12/ 2*Pi*f*(0.21*10-6)
1*10-3 = 12/ 2*Pi*f*(0.21*10-6)
f = 9.094*109 Hz = 9.09 GHz
(2) We know that XL = 2Pi*f*L = 2*Pi*60*240*10-3 = 90.478 ohm
XC = 1/2Pi*fC = 1/(2*Pi*60*1.4*10-6) = 1894.707 ohm
Resistance (R) = 9.9*1000 = 9900 ohm
We know that the phase angle is = tan-1(XL -XC /R)= -10.328 degree = -10.4 degree
(3) XL = XC
2*Pi*fL = 1/(2Pi*fC)
f2 = (1/2*Pi)2*(1/LC) = (1/2Pi)2(1/35*10-3*35*10-6)
f = 143.797 Hz = 1.4*102 Hz
(4) Resonanace frequency is the one at which the reactance of the inductor and the capacitor becomes equal therefore
XL = XC
Which we have calculated in the above part .
therefore fResonance = 1.4*102 Hz

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