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An Am radio antenna picks up a 1000.0 kHz signal witha peak voltage of 6.0 mV. T

ID: 1988639 • Letter: A

Question

An Am radio antenna picks up a 1000.0 kHz signal witha peak voltage of 6.0 mV. The tuning circuit consists of a 70.0 µH inductor in series with a variable capacitor. The inductor coil has a resistance of 0.20 O, and the resistance of the rest of the signal is negligible.
To what value should the capacitor be tuned to listen to this radio station?

What is the peak current through the circuit at resonance?

A stronger station at 1050.0 kHz produces a 12.0 mV antenna signal. What is the current at this frequency when the radio is tuned to 1000.0 kHz? Think about how this current compares to the current you found in part b.

Explanation / Answer

Given Frequency of antenna f = 1000 kHz                                      = ( 1000 kHz)( 1000 Hz / 1 kHz )                                      = 10 6 Hz Maximum voltage V = 6 mV                                   = ( 6 mV ) ( 0.001 V / mV )                                   = 0.006 V Inductance of the inductor L = 70 H                                              = ( 70 H ) ( 10-6 H / 1 H )                                              = 70 *10-6 H Resistance of the resistor R = 20 a) resonance frequency                 f = 1 / 2 LC                LC = 1 / 2f                         = 1 / 2 ( 106 Hz )                    LC = 0.025*10-12 Capacitance of the capacitor is                      C = (0.025*10-12) / ( 70 *10-6 H )                          = 3.57*10-10 F ___________________________________________________ b) A t resonance frequency impedance is equal to resistance .     So the peak current through the circuit is                      I = V / R                         = 0.006 V / 20                         = 0.0003 A ___________________________________________________ ___________________________________________________ c) Frequency of antenna f = 1050 kHz                                      = ( 1050 kHz)( 1000 Hz / 1 kHz )                                      = 1.05*10 6 Hz Maximum voltage V =  12 mV                                   = ( 6 mV ) ( 0.001 V / mV )                                   = 0.012 V Inductive reactance            X L = 2 fL                   = 2 (1.05*10 6 Hz ) ( 70 *10-6 H )                   = 461.81 Capacitive reactance           X C = 1 / 2fC                  = 1 / 2 ((1.05*10 6 Hz )( 3.57*10-10 F )                  = 424.5 Impedance of the circuit is            Z = ( R 2 + ( X L - X C ) 2 ) 1/2                = ( 202 + ( 461.81 - 424.5 ) 2 ) 1/2                = 42.332 Maximum current           I = V / Z              = 0.012 / 42.332              = 0.0002834 A Frequency of antenna f = 1050 kHz                                      = ( 1050 kHz)( 1000 Hz / 1 kHz )                                      = 1.05*10 6 Hz Maximum voltage V =  12 mV                                   = ( 6 mV ) ( 0.001 V / mV )                                   = 0.012 V Inductive reactance            X L = 2 fL                   = 2 (1.05*10 6 Hz ) ( 70 *10-6 H )                   = 461.81 Capacitive reactance           X C = 1 / 2fC                  = 1 / 2 ((1.05*10 6 Hz )( 3.57*10-10 F )                  = 424.5 Impedance of the circuit is            Z = ( R 2 + ( X L - X C ) 2 ) 1/2                = ( 202 + ( 461.81 - 424.5 ) 2 ) 1/2                = 42.332 Maximum current           I = V / Z              = 0.012 / 42.332              = 0.0002834 A
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