Question

A voltage of peak value $283\text{V}$ and varying frequency is applied to a series $L-C-R$ combination in which $R=3\Omega ;L=25\text{mH}$ and $C=400\mu F$. Then, the frequency (in $Hz$) of the source at which maximum power is dissipated in the above, is

• A. $51.6$
• B. $50.7$
• C. $51.1$
• D. $50.3$

Answer 1

The correct option is D $50.3$

Given
$V_0=283\text{V}R=3\Omega$
$L=25\times {10}^{-3}\text{H}C=400\times {10}^{-6}\text{F}$
For maximum power $X_L=X_C$
$\omega C=\frac{1}{\omega L}\Rightarrow {\omega }^2=\frac{1}{LC}$
$\omega =\frac{1}{\sqrt{LC}}$
$f=\frac{\omega }{2\pi }=\frac{1}{2\pi \sqrt{LC}}$
$=\frac{1}{2\times 3.14}\times \frac{1}{\sqrt{25\times {10}^{-3}\times 400\times {10}^{-6}}}=50.3$