Question

The antenna current of an $AM$ transmitter is $8\,A$ when only the carrier is sent, but it increases to $8.93\,A$ when the carrier is modulated by a single sine wave. Find the percentage modulation.

• A. 60.1 %
• B. 70.1 %
• C. 80.1 %
• D. 50.1 %

Answer 1

Answer: (B)

The power dissipated in any circuit is a function of the square of voltage across the circuit and the effective resistance of the circuit. Equation of AM wave reveals that it has three components of amplitude $E_{c}, m E_{c} / 2$ and $m E_{c} / 2 .$ Clearly, power output must be distributed among these components.
Carrier Power, $P_{C}=\frac{\left(E_{c} / \sqrt{2}\right)^{2}}{R}=\frac{E_{c}^{2}}{2 R}$
Total power of side bands,
$P_{S}=\frac{\left(m E_{c} / 2 \sqrt{2}\right)^{2}}{R}+\frac{\left(m E_{c} / 2 \sqrt{2}\right)^{2}}{R} $
$=\frac{m^{2} E_{c}^{2}}{4 R}$
$\therefore \frac{P_{S}}{P_{C}}=\frac{1}{2} m^{2}$
and $P_{T}=P_{C}+P_{S}=P_{C}\left(1+\frac{m^{2}}{2}\right)$
$\therefore \frac{P_{T}}{P_{C}}=1+\frac{m^{2}}{2} $
or $\left(\frac{I_{T}}{I_{C}}\right)^{2}=1+\frac{m^{2}}{2}$
Given that, $I_{T}=8.93 A , I_{C}=8 A , m=$ ?
$\therefore \left(\frac{8.93}{8}\right)^{2}=1+\frac{m^{2}}{2}$
or $1.246=1+\frac{m^{2}}{2}$
or $\frac{m^{2}}{2}=0.246$
or $m=\sqrt{2 \times 0.246}=0.701$
$=70.1 \%$
The situation is summarised in figure. $B C=A D=3 \,cm , A B=D C=4 cm ,$ so $A C=5 \,cm$