Question

A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effective value of 110 V is applied to the primary coil. At the secondary coil, an alternating voltage with an effective value of 550 V is obtained. A current of 29 A is supplied to the primary coil of the transformer. Calculate the maximum effective current in the secondary coil. The maximum effective current in the secondary coil is A.

Answer 1

Answer:

$I_s=5.8A$

Explanation:

Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:

$P_p=P_s$

So:

$V_p*I_p=V_s*I_s$

Where:

$V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil$

Solving for $I_s$

$I_s=\frac{V_p*I_p}{V_s}$

Replacing the data provided:

$I_s=\frac{110*29}{550} =5.8A$