A step-down transformer has a N_{S} to N_{P} | vedclass

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(A) Given: Ratio of secondary turns to primary turns $\frac{N_{S}}{N_{P}} = \frac{1}{20}$.Secondary voltage $V_{S} = 8 \, V$.Secondary resistance $R_{

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(A) Given: Ratio of secondary turns to primary turns $\frac{N_{S}}{N_{P}} = \frac{1}{20}$.Secondary voltage $V_{S} = 8 \, V$.Secondary resistance $R_{S} = 0.4 \, \Omega$.First, calculate the secondary current $I_{S}$ using Ohm's law: $I_{S} = \frac{V_{S}}{R_{S}} = \frac{8 \, V}{0.4 \, \Omega} = 20 \, A$.For an ideal transformer, the relationship between currents and turns is given by $\frac{I_{P}}{I_{S}} = \frac{N_{S}}{N_{P}}$.Substituting the values: $I_{P} = I_{S} 	imes \frac{N_{S}}{N_{P}} = 20 \, A 	imes \frac{1}{20} = 1 \, A$.Therefore, the primary current is $1 \, A$.

$A$ step-down transformer has a $N_{S}$ to $N_{P}$ ratio of $1: 20$. If $8 \, V$ are developed across a $0.4 \, \Omega$ secondary load, the primary current will be: (in $ \, A$)

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