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NEET UG PHYSICS MCQ: ALTERNATING CURRENT

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Application +4 | -1
In an ideal transformer, the voltage is stepped down from 11 kV to 220 V. If the primary current be 100 A, the current in the secondary should be
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Correct option: C
0.5 kA
Option A (5 kA) is correct because in an ideal transformer, the power is conserved, so VPIP=VSISV_P I_P = V_S I_S, which gives IS=(VPIP)/VS=(11000∗100)/220=5000A=5kAI_S = (V_P I_P) / V_S = (11000 * 100) / 220 = 5000 A = 5 kA.

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