0 votes 0 votes The output voltage of the ideal transformer with the polarities and dots shown in the figure is given by $NV_i\sin \omega t$ $-NV_i\sin \omega t$ $\frac{1}{N}V_i\sin \omega t$ $-\frac{1}{N}V_i\sin \omega t$ Others gate2015-in + – Milicevic3306 asked Mar 26, 2018 • edited Nov 23, 2020 by soujanyareddy13 Milicevic3306 7.9k points answer See all 0 reply
