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(D) For an ideal (lossless) transformer,the power input equals the power output,which implies $V_p i_p = V_s i_s$.Since the voltage ratio is proportio

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$10$

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(D) For an ideal (lossless) transformer,the power input equals the power output,which implies $V_p i_p = V_s i_s$.Since the voltage ratio is proportional to the turns ratio,we have $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.For a lossless transformer,the current ratio is inversely proportional to the turns ratio: $\frac{i_s}{i_p} = \frac{N_p}{N_s}$.Given: $i_p = 2 \ A$,$N_p = 100$,$N_s = 20$.Substituting the values: $i_s = i_p 	imes \frac{N_p}{N_s} = 2 	imes \frac{100}{20}$.$i_s = 2 	imes 5 = 10 \ A$.

In a lossless transformer,an alternating current of $2 \ A$ is flowing in the primary coil. The number of turns in the primary and secondary coils are $100$ and $20$ respectively. The value of the current in the secondary coil is.......$A$

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