Question

A transformer has a primary coil with 106 turns and a secondary coil of 340 turns. The AC voltage across the primary coil has a maximum of 128 V and the AC current through the primary coil has a maximum of 6 A. What are the maximum values of the voltage and current for the secondary coil?

Answer 1

To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.

From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

Where

$V_p =$  input voltage on the primary coil.

$V_s=$input voltage on the secondary coil.

$N_p=$  number of turns of wire on the primary coil.

$N_s =$ number of turns of wire on the secondary  coil.

Replacing our values we have:

$V_p = 128V$

$N_p = 106$

$N_s = 340$

Replacing,

$\frac{V_s}{128} = \frac{340}{106}$

$V_s = 410.56V$

From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:

$V_s I_s = V_p I_p$

Where

$I_p$= Current Primary Coil

$I_s$ = Current secundary Coil

Therefore:

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

$I_s = \frac{(128)(6)}{410.56}$

$I_s = 1.87 A$

Therefore the maximum values for the secondary coil of the voltage is 410.56V and Current is 1.87A