Exams › JEE Main › Physics
The modulation index for amplitude modulation, expressed using the maximum and minimum values of the envelope, is given by which relation?
- mₐ = (E_max + E_min) / E_min
- mₐ = (E_max - E_min) / E_max
- mₐ = (E_max - E_min) / (E_max + E_min)
- mₐ = (E_max + E_min) / (E_max - E_min)
Correct answer: mₐ = (E_max - E_min) / (E_max + E_min)
Solution
The modulation index for amplitude modulation is defined as the ratio of the difference between the maximum and minimum envelope values to the sum of those values, which effectively measures the extent of modulation in the signal.
Related JEE Main Physics questions
- An AC source of angular frequency ω is connected across a series combination of a resistor r and a capacitor C. The circuit current is I. If the source frequency is reduced to ω/3 while the applied voltage amplitude remains unchanged, the current becomes I/2. The ratio of capacitive reactance to resistance at the original frequency ω is
- A series LCR circuit has a resistor of 200 Ω and is connected to a 220 V, 50 Hz AC supply. If the capacitor is removed first and then the inductor is removed, the voltage across the resistor in the circuit is found to lead the supply voltage by 30°. Also, when the inductor is removed, the current is observed to lead the voltage by 30°. The power dissipated in the circuit is
- A radio receiver’s tuned circuit contains a 50 Ω resistance, a 10 mH inductor, and a variable capacitor. If a 1 MHz radio signal induces a potential difference of 0.1 mV, the capacitor value required for resonance is
- In a circuit, the instantaneous current and voltage are expressed by
i = (1/√2) sin(100πt) A
e = (1/√2) sin(100πt + π/3) V
What is the average power, in watts, dissipated in the circuit?
- In a series L-C circuit, the current reaches its peak when the inductance is 0.5 H and the capacitance is 8 μF. What is the angular frequency of the applied alternating voltage?
- A coil has a resistance of 30 ohm and an inductive reactance of 20 ohm when the frequency is 50 Hz. If a 200 V AC supply of frequency 100 Hz is applied across the coil, the current through it will be
⚔️ Practice JEE Main Physics free + battle 1v1 →