Given that \(A_c=5V\), Resistance \(R=20\mathrm{\Omega}\)
Carrier frequency \(f_c=\frac{6\times{10}^8}{2\pi}=95.45MHz\)
Modulating signal frequency \(f_m=\frac{1000}{2\pi}=159.10Hz\)
Modulation index \(\beta=5\)
Maximum deviation \(∆f=\beta f_{m}=5\times159.10=795.45Hz\)
Power dissipated by the load \(20\mathrm{\Omega}\) is
\(P=\frac{\frac{1}{2}{A_c}^2}{R}=\frac{\frac{1}{2}\times25}{20}=0.625watts\)