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(B) Given:Primary voltage,$V_p = 220\,V$Secondary voltage,$V_s = 11\,kV = 11000\,V$Power transmitted,$P = 4.4\,kW = 4400\,W$Since the transformer is $

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$0.4$

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(B) Given:Primary voltage,$V_p = 220\,V$Secondary voltage,$V_s = 11\,kV = 11000\,V$Power transmitted,$P = 4.4\,kW = 4400\,W$Since the transformer is $100\%$ efficient,the power in the secondary coil is equal to the power in the primary coil,i.e.,$P_s = P_p = 4400\,W$.The current in the secondary coil $(I_s)$ is given by the formula:$I_s = \frac{P_s}{V_s}$Substituting the values:$I_s = \frac{4400\,W}{11000\,V} = 0.4\,A$Therefore,the current rating of the secondary is $0.4\,A$.

$A$ power transformer is used to step up an alternating $e.m.f.$ of $220\,V$ to $11\,kV$ to transmit $4.4\,kW$ of power. If the primary coil has $1000$ turns,what is the current rating of the secondary (in $,A$)? Assume $100\%$ efficiency for the transformer.

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