Q.Derive the current and power relations for AM wave in terms of modulation index.

The total power of an AM wave equals the carrier power plus the power in the two side bands, giving $P_t=P_c(1+m_a^{2}/2)$. Since power is proportional to the square of the antenna current, the current relation follows as $I_t=I_c\sqrt{1+m_a^{2}/2}$.

[!ANSWER]

Power: $P_t=P_c\left(1+\dfrac{m_a^{2}}{2}\right)$. Current: $I_t=I_c\sqrt{1+\dfrac{m_a^{2}}{2}}$.

[!TLDR]

For an AM wave $P_t=P_c(1+m_a^2/2)$ and, since $P\propto I^2$, $I_t=I_c\sqrt{1+m_a^2/2}$.

Power relation:

The AM wave has a carrier of amplitude $E_c$ and two side bands each of amplitude $m_aE_c/2$. The power delivered to a resistance $R$ by a sinusoid of peak value $E$ is $P=E^{2}/2R$.

Carrier power:

Power in each side band:

The two side bands together carry:

Total power:

Current relation:

Let $I_c$ be the rms antenna (carrier) current when unmodulated and $I_t$ the rms current when modulated, both flowing in the same antenna resistance $R$. Since power is proportional to the square of the current in the same resistance,

But $P_t/P_c = 1+m_a^{2}/2$, so

These relations are used in transmitter measurements: by reading the unmodulated and modulated antenna currents (or powers) the depth of modulation $m_a$ can be calculated.

[!ANSWER]

The power relation is $P_t=P_c\left(1+\dfrac{m_a^{2}}{2}\right)$ and the current relation is $I_t=I_c\sqrt{1+\dfrac{m_a^{2}}{2}}$, where $P_c, I_c$ are the unmodulated carrier power and current and $P_t, I_t$ the corresponding modulated (total) values.