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The number of turns in the primary and secondary coils of an ideal transformer are 140 and 280, respectively. If the current through the primary coils is 4 A, what will be the current in the secondary coil? |
Answer» In an ideal transformer, the powers of the secondary and primary coild are equal, i.e., `V_sI_s=V_pI_p` `therefore I_s=I_p.p/V_s=I_P.N_P/N_s=4times140/280=2A`. |
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