Question

$3-\varphi$ full converter feeds power to resistive load of $10\Omega$. If the firing angle is ${30}^{\circ }$, the load takes 8 kW . The magnitude of per phase input supply voltage is____ V.

• A. 137.36

Answer 1

The correct option is A 137.36

Given, $\alpha ={30}^{\circ }$

As the connected load is resistive, output current waveform is same as output voltage waveform.

If firing angle $\alpha \le {60}^{\circ }$

The output voltage and current will be continuous

$\therefore$ Rms output voltage

$(V_0{)}_{rms}={\left[\frac{3}{\pi }{\int }_{\frac{\pi }{3}+\alpha }^{\frac{2\pi }{3}+\alpha }{V}_{ml}^2{\sin }^2\omega td\right(\omega t\left)\right]}^{1/2}$

$(V_0{)}_{rms}^2={\left[\frac{3V_{ml}^2}{2\pi }{\int }_{\frac{\pi }{3}+\alpha }^{\frac{2\pi }{3}+\alpha }\right(1-\cos 2\omega t\left)d\omega t\right]}^{1/2}$

$(V_0{)}_{rms}=V_{ml}\sqrt{\frac{3}{2\pi }}{[\frac{\pi }{3}+\frac{\sqrt{3}}{2}\cos 2\alpha ]}^{1/2}$

Given , $\frac{V_{0,rms}^2}{R}=8000W$

$V_{ml}=\sqrt{2}{V}_s$

$2V_s^2\frac{3}{2\pi }[\frac{\pi }{3}+\frac{\sqrt{3}}{2}\cos {60}^{\circ }]=8000\times 10$

$V_s^2\frac{6}{2\pi }[1.0471+0.4330]=8000\times 10$

$V_s^2\frac{6}{2\pi }\left[1.4801\right]=80000$

$V_s^2=56601.448$

$V_s=237.91$

$\therefore \text{per phase voltage}=\frac{V_s}{\sqrt{3}}=\frac{237.91}{\sqrt{3}}=137.36V$