Question

An AM radio antenna pi

ks up a 1000 kHz signal with a peak voltage of 5.0 mV. The tuning

ir

uit


onsists of a 60 µH indu

tor in series with a variable

apa

itor. The indu

tor

oil has a resistan

e of

0.25 Ω, and the resistan

e of the rest of the

ir

uit is negligible.

(a) To what value should the

apa

itor be tuned to listen to this radio station? (Note that the


apa

itor needs to be tuned to where it is in resonan

e with the indu

tor).

(b) Cal

ulate the peak

urrent through the

ir

uit at resonan

e.

(

) A stronger station at 1050 kHz produ

es 10 mV antenna signal. Cal

ulate the

urrent at this

frequen

y and the

orresponding phase angle when the radio is tuned to 1000 kHz.

Answer 1

Given that $R = 0.25\Omega$, $L = 60 \mu H = 6\cdot 10^{-5}H$, $f = 1000 kHz$.

a) Find $C$ in the series RLC resonant circuit. The frequency of the resonant ciruit is given by:

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}$

Find $C$ from this expression:

$C = \dfrac{4L}{16\pi^2 f^2L^2 + R^2} = \dfrac{4\cdot 6\cdot 10^{-5}}{16\pi^2 \cdot 10^{12}\cdot 36\cdot 10^{-10} + 0.0625} = 4,22\cdot 10^{-10} F = 0.422 nF$.

b) Assume that antenna signal is $U = 10 mV$. Then the peak current at $f = 1 MHz$ is:

$I = \dfrac{U}{Z}$ , where $Z$ is the module of impedance.

For the series RLC circuit:

$Z = \sqrt{R^2 + (2\pi fL - \frac{1}{2\pi fC})^2 } = \sqrt{0.25^2 + (2\pi 10^6\cdot 6\cdot 10^{-5} - \frac{1}{2\pi 10^6\cdot 0.422\cdot 10^{-9}})^2} = 0.29 \Omega$.

Thus, the current is:

$I = \dfrac{10\cdot 10^{-3}}{0.29} = 0.03448A= 34.5mA$.

c)According to the formula for the capacitance from the part a), the $C$ for $f = 1050 kHz$ will be:

$C = \dfrac{4L}{16\pi^2 f^2L^2 + R^2} = \dfrac{4\cdot 6\cdot 10^{-5}}{16\pi^2\cdot 1050^2 \cdot 10^{6}\cdot 36\cdot 10^{-10} + 0.0625} = 3.83\cdot 10^{-10} F = 0.383 nF$

The impedance then is:

$Z = \sqrt{R^2 + (2\pi fL - \frac{1}{2\pi fC})^2 } = \sqrt{0.25^2 + (2\pi\cdot 1050\cdot 10^3\cdot 6\cdot 10^{-5} - \frac{1}{2\pi \cdot 1050\cdot 10^3\cdot 0.383\cdot 10^{-9}})^2} = 0.26 \Omega$

The current will be:

$I = \dfrac{10\cdot 10^{-3}}{0.26} = 0.03870A= 38.7mA$

d) The phase angle is given by:

$\varphi = \arctan(\dfrac{2\pi fL - \frac{1}{2\pi fC}}{R}) = \arctan(\dfrac{2\pi 10^6\cdot 6\cdot 10^{-5} - \frac{1}{2\pi 10^6\cdot 0.422\cdot 10^{-9}}}{0.25}) = 31.5\degree C.$