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An ideal transformer converts 200 V  AC to 3.3 kV  AC to transmit a power of 4.4 kW .If primary coil has 600 turns, then alternating current in the secondary coil is 

Answer:

Explanation:

 Given, input voltage (V_p) =220 V

 Output voltage (V_s) = 3.3 x 10³ V

 Power (P) = 4.4 kW = 4.4 x 10³ W

 Number of turns in primary coil = 600

 We know that,

   $P= V_{p}\times I_{p} \Rightarrow I_{p}=\frac{P}{V_{p}} or I_{p}=\frac{4.4\times 10^{3}}{220}=20A$

 For an ideal transformer,

$\frac{V_{s}}{V_{p}}=\frac{I_{p}}{I_{s}}$

$I_{s}=I_{p}\times\frac{V_{p}}{V_{s}}$

$I_{s}=\frac{20\times220}{3.3 \times 10^{3}}$

 or  $I_{s}=\frac{44}{33}=\frac{4}{3}A$