An amplitude modulated voltage signal \(x(t)\) drives a load of \(1\,\Omega\). \[
x(t) = K\cos(300\pi t) + L\cos(240\pi t) + L\cos(360\pi t)
\] If the efficiency is \(60\%\) and the carrier power is \(50\) W, the value of \(L\) is ______ V.
Show Hint
Split the terms into one carrier and two equal sidebands, use \(P_c=K^2/2R\) and \(\eta=P_{sb}/(P_c+P_{sb})\) to get \(P_{sb}\), then \(P_{sb}=L^2\) since \(R=1\,\Omega\).
Step 1: Recast the problem using the modulation index.
For a single-tone AM signal $K[1+\mu\cos(\omega_m t)]\cos(\omega_c t)$, expanding the product gives a carrier term $K\cos(\omega_c t)$ and two sideband terms each of amplitude $\dfrac{K\mu}{2}$. Comparing with the given signal, the sideband amplitude here is $L$, so
\[
L = \frac{K\mu}{2}
\] Step 2: Recall the standard AM power-efficiency formula in terms of $\mu$.
For single-tone AM, the fraction of total power carried by the sidebands is
\[
\eta = \frac{\mu^2}{2+\mu^2}
\] Step 3: Solve for the modulation index.
\[
0.6 = \frac{\mu^2}{2+\mu^2}
\]
\[
0.6(2+\mu^2) = \mu^2
\]
\[
1.2 + 0.6\mu^2 = \mu^2
\]
\[
1.2 = 0.4\mu^2 \ \Rightarrow\ \mu^2 = 3 \ \Rightarrow\ \mu=\sqrt3\approx1.732
\] Step 4: Find K from the carrier power.
\[
P_c=\frac{K^2}{2R}=50,\quad R=1\,\Omega \ \Rightarrow\ K^2=100 \ \Rightarrow\ K=10\ \text{V}
\] Step 5: Combine to get L.
\[
L = \frac{K\mu}{2} = \frac{10\times1.732}{2} = 8.66\ \text{V}
\]
This matches the power-balance method exactly, confirming the result through the modulation-index route instead of working with $P_{sb}$ directly. The other options do not correspond to any consistent value of $\mu$ between 0 and the point where $\eta$ would exceed $100\%$, so they can be discarded once $\mu$ is pinned down uniquely by $\eta=0.6$.
\[
\boxed{L\approx8.66\ \text{V}}
\]
