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Question

(A) For an ideal transformer,the relationship between voltage,turns,and current is given by: $\frac{E_{S}}{E_{P}} = \frac{n_{S}}{n_{P}} = \frac{I_{P}}

Vedclass · Accepted Answer

$240$ $V$,$5$ $A$

Vedclass · Answer

(A) For an ideal transformer,the relationship between voltage,turns,and current is given by: $\frac{E_{S}}{E_{P}} = \frac{n_{S}}{n_{P}} = \frac{I_{P}}{I_{S}}$.First,calculate the secondary voltage $(E_{S})$:$E_{S} = E_{P} 	imes \frac{n_{S}}{n_{P}} = 120 	imes \frac{200}{100} = 240$ $V$.Next,calculate the secondary current $(I_{S})$:$I_{S} = I_{P} 	imes \frac{n_{P}}{n_{S}} = 10 	imes \frac{100}{200} = 5$ $A$.Therefore,the voltage and current in the secondary are $240$ $V$ and $5$ $A$ respectively.

The primary winding of a transformer has $100$ turns and its secondary winding has $200$ turns. The primary is connected to an $AC$ supply of $120$ $V$ and the current flowing in it is $10$ $A$. The voltage and the current in the secondary are:

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