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=(840*0.3)/(56) \\ =4.5\text{ A} \end{gathered}`

The secondary voltage can be calculated by the formula

`\begin{gathered} V_s=(V_pI_p)/(I_s) \\ =(110*0.3)/(4.5) \\ =7.33\text{ V} \end{gathered}`

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