Question

A single phase full bridge inverter has an RLC load with $R=10\Omega ,L=31.5mH$ and $C=122\mu F$. The inverter frequency is $f_0=60Hz$ and dc input voltage is $V_s=220V$. If considering upto ninth harmonic, the rms value of load current is $13.02A,$ then the % THD of the load current is

• A. 17.91%
• B. 19.24%
• C. 18.62%
• D. 17.12%

Answer 1

The correct option is C 18.62%

The rms value of fundamental voltage component is
$V_{01}=\frac{4V_s}{\pi \sqrt{2}}=\frac{4\times 220}{\pi \sqrt{2}}=198.07V$
$|Z|=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$
$=\sqrt{{10}^2{(2\pi \times 60\times 3.15\times {10}^{-3}-\frac{1}{2\pi \times 60\times 112\times {10}^{-6}})}^2}$
$|Z|=15.47\Omega$
The rms value of fundamental load current is,
$i_{01\left(rms\right)}=\frac{V_{01}\left(rms\right)}{|Z|}=\frac{198.07}{15.47}=12.8A$
% THD of load current $=\frac{\sqrt{I_{rms}^2-I_{01rms}^2}}{I_{01rms}}\times 100\%$

$=\frac{\sqrt{(13.02{)}^2-(12.8{)}^2}}{12.8}\times 100\%=18.62\%$