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A transformer with efficiency `80%` works at `4 kW` and `100 V`. If the secondary voltage is `200 V`, then the primary and secondary currents are respectivelyA. 16A, 40AB. 40A, 16AC. 30A, 45AD. 50A, 30A |
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Answer» Correct Answer - B `P_(i) = 4 xx 10^(3) W` `:. I_(P) = (P_i)/(v) = (4 xx 10^3)/(100) = 40 A` (Primary) `eta = (P_o)/(P_i) :. 8/10 = (P_o)/(4 xx 10^3) :. P_(o) = 32 xx 10^(2)` `:. I_(S) = (P_o)/(V_S) = (32 xx 10^(2))/(200) = 16 A` (Secondary). |
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