**A 3 phase dual converter is operated front the secondary of a delta star connected transformer of 220 volts, 50 Hz supply. If the load resistance is 10 Ω, the circulating inductance is 7.5 mH and \({\alpha _1} = {50^0}\), find, (i) Peak circulating current (ii) Peak current of converter 1 (AMIE Summer 2023)**

**Solution**

Secondary phase voltage

\({V_{ph}} = \frac{{220}}{{\sqrt 3 }} = 127\,V\)

Peak value of phase voltage

\(\begin{array}{l}{V_m} = \sqrt 2 {V_{ph}} = \sqrt 2 \,x\,127 = 179.63\,V\\\omega = 2\pi f = 2\pi \,x\,50 = 314.16\,rad/s\end{array}\)

*Peak circulating current*

\({i_{cir}}(t) = \frac{{3{V_m}}}{{\omega {L_r}}}\left[ {\sin \left( {\omega t - \frac{\pi }{6}} \right) - \sin {\alpha _1}} \right]\)

This current will be maximum when

\(\sin \left( {\omega t - \frac{\pi }{6}} \right) = 1 \Rightarrow \omega t = \frac{{2\pi }}{3}\)

\({i_{cir(peak)}} = \frac{{3x179.63\,V}}{{314.16(7.5\,x\,{{10}^{ - 3}})}}\left[ {\sin \left( {\frac{{2\pi }}{3} - \frac{{2\pi }}{6}} \right)} \right] = 53.5\,A\)

*Peak current of converter 1*

Peak load current

\( = \frac{{2{V_m}}}{R} = \frac{{3(179.63)}}{{10}} = 53.89\,A\)

\(\begin{array}{l}{i_{peak,conv1}} = peak\,load\,current + {\rm{ }}peak\,circulating\,current\\{\rm{ = 53}}{\rm{.5}}\,{\rm{ + }}\,{\rm{53}}{\rm{.89}}\,{\rm{ = }}\,{\rm{107}}{\rm{.39}}\,{\rm{A}}\end{array}\)

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