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:

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