Question

A single phase full converter charges a battery which offers a constant voltage $E=12V$. A resistor R is inserted to limit the battery charging current to $6A$. If supply voltage is $40V,50Hz$, then the value of $R=$ ________ $\Omega$.

Answer 1

Average value of load current is, $\begin{array}{cc}{I}_0& =\frac{1}{\pi R}{\int }_{\theta }^{-\pi -\theta }[V_m\sin \omega t-(E+V_d\left)\right]d\omega t\\ Asweknow,V_m\sin \omega t& =E+V_d\\ 40\sqrt{2}\sin \theta & =12+1\\ \theta & ={\sin }^{-1}\left(\frac{13}{40\sqrt{2}}\right)={13.28}^{\circ }\\ 6& =\frac{1}{\pi R}{\int }_{{13.28}^{\circ }}^{{166.72}^{\circ }}(40\sqrt{2}\sin \omega t-13d\omega t)\\ R& =\frac{1}{6\pi }[-40\sqrt{2}(\cos {166.72}^{\circ }-\cos {13.28}^{\circ })-13({166.72}^{\circ }-{13.28}^{\circ })\frac{\pi }{{180}^{\circ }}]\\ R& =3.99\Omega \end{array}$