Question

If the length of a linear antenna is increased by $60%$ and the wavelength of the signal is decreased by $20%$, then the percentage increase in the effective power radiated by the antenna is:

• A. $50$
• B. $250$
• C. $300$
• D. $150$

Hint:

If the final value becomes four times the original value, the increase is not $400%$. Since one original value already existed, the increase is $(4-1)\times100=300%$.

Answer 1

The Correct Option is C

Concept: The radiated power of a linear antenna is proportional to \[ P\propto\left(\frac{l}{\lambda}\right)^2 \] where $l$ is the antenna length and $\lambda$ is the wavelength. Therefore, \[ \frac{P_2}{P_1} = \left(\frac{l_2}{l_1}\cdot\frac{\lambda_1}{\lambda_2}\right)^2 \]

Step 1: Write the modified quantities Length increases by $60%$: \[ l_2=1.6l_1 \] Wavelength decreases by $20%$: \[ \lambda_2=0.8\lambda_1 \]

Step 2: Find the power ratio \[ \frac{P_2}{P_1} = \left(\frac{1.6}{0.8}\right)^2 \] \[ =(2)^2 \] \[ =4 \] Thus, \[ P_2=4P_1 \]

Step 3: Calculate percentage increase \[ %\text{ Increase} = \left(\frac{P_2-P_1}{P_1}\right)\times100 \] \[ =(4-1)\times100 \] \[ =300% \] Therefore, \[ \boxed{300%} \]