Question

The antenna current of an AM transmitter is 8A when only the carrier is sent, but it increases to 8.93A 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

Hint: In amplitude modulation the modulating wave generally the audio waves are superimposed on a carrier wave generally the radio waves. The frequency of the modulated and carrier wave is the same but the amplitude varies as in accordance with the instantaneous amplitude of the modulating wave. In general, the use of transistors in this field is done by referring to the power amplification, etc.

Formula used:
Formula relating power of modulated and carrier wave which is given as:
\[{P_m} = {P_c}(1 + \dfrac{{{m^2}}}{2})\]
Where, ${P_m}$is the power of modulated wave
$Pc$is the power of carrier wave
m is the modulation index

Complete step by step answer:
 Modulation is the most essential thing in today’s world. Without modulation we would not be able to access long distance phone calls, internet etc. Modulation is the superimposing of a modulating wave onto a carrier wave and transmitting it which is again demodulated at the receiving station. The modulating wave cannot be transmitted as the frequency/amplitude is high which requires more energy or an antenna of the range of kilo-meters. Hence, we use these carrier waves which are easy and require less energy to get transmitted.
Modulations are of three types- Amplitude, Frequency and Phase out of which frequency modulation is the most common. In amplitude modulation, the frequency of the modulated and carrier wave is the same but the amplitude varies as in accordance with the
instantaneous amplitude of the modulating wave. The power formula is given as
\[{P_m} = {P_c}(1 - \dfrac{{{m^2}}}{2})\]
But power can be written in terms of current as
\[P = {I^2}R\]
Substituting the above expression in the former formula we have (obviously, the resistance is the same),
\[
I_m^2R = I_c^2R(1 + \dfrac{{{m^2}}}{2}) \\
\dfrac{{I_m^2 {R}}}{{I_c^2 {R}}} = (1 + \dfrac{{{m^2}}}{2}) \\
m = \sqrt {2(\dfrac{{I_m^2}}{{I_c^2}} - 1)} \\
\]
On substituting the data, we get,
\[
m = \sqrt {2(\dfrac{{{{8.93}^2}}}{{{{8.00}^2}}} - 1)} = \sqrt {2 \times 0.246} \\
\Rightarrow m = 0.701 \\
or \\
m = 70.1\% \\
 \]
The modulating index is 0.701 and percentage modulation is 70.1
The correct answer is option B.

Note: The power formula is a very common expression and can be found in almost every book. Hence, we can either convert in current or voltage expressions. Current has already been done. For voltage substitute $P = \dfrac{{{V^2}}}{R}$. Therefore, the expression becomes \[m = \sqrt {2(\dfrac{{V_m^2}}{{V_c^2}} - 1)} \].