Question

A transformer has 840 primary and 56 secondary windings. The primary coil is connected to a 110 V AC power supply which delivers 0.30 A of current to the transformer.a. Find the secondary voltageb. Find the secondary current

Answer 1

Given that the number of turns in the primary coil is

$N_p=\text{ 840 }$

The number of turns in the secondary coil is

$N_s=\text{ 56}$

The voltage provided to the primary coil is

$V_p=110\text{ V}$

The current delivered to the transformer is

$I_p=\text{ 0.30 A}$

We have to find the secondary voltage and secondary current.

Let the secondary current be denoted by

$I_s$

Let the secondary voltage be denoted by

$V_s$

The secondary current can be calculated by the formula

$I_s=\frac{N_pI_p_{}}{N_s}$

Substituting the values, the secondary current will be

$\begin{gathered} I_s=\frac{840\times0.3}{56} \\ =4.5\text{ A} \end{gathered}$

The secondary voltage can be calculated by the formula

$\begin{gathered} V_s=\frac{V_pI_p}{I_s} \\ =\frac{110\times0.3}{4.5} \\ =7.33\text{ V} \end{gathered}$

Thus, the secondary voltage is 7.33 V and the secondary current is 4.5 A