Question

A transformer has $500\mathrm{turns}$ in its primary and $1000\mathrm{turns}$ in its secondary winding. The primary voltage is $200\mathrm{V}$ and the load in the secondary is $100\mathrm{ohm}$. Calculate the current in the primary, assuming it to be an ideal transformer.

Answer 1

Step 1: Given information

In its primary winding, a transformer has $500\mathrm{turns}$.

In its secondary winding, a transformer has $1000\mathrm{turns}$.

The primary voltage is $200\mathrm{V}$.

The load resistance in the secondary winding is $100\mathrm{ohm}$.

Step 2: To find

We have to find the current in the primary winding.

Step 3: Calculation

We know the relation,

$\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}}=\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}}$

or,

${\mathrm{V}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{p}}\times \frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}}$

Here,

V_p is the primary voltage.

V_s is the secondary voltage.

N_s is the secondary turns.

N_p is the primary turns.

By substituting the given values, we get

$$\begin{gathered} {\mathrm{V}}_{\mathrm{s}}=200\times \frac{1000}{500} \\ =400\mathrm{V} \end{gathered}$$

The current in the secondary winding will be:

${\mathrm{i}}_{\mathrm{s}}=\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{R}}_{\mathrm{s}}}$

Here, R_s is the load resistance.

By substituting the values,

$$\begin{gathered} =\frac{400}{100} \\ =4\mathrm{A} \end{gathered}$$

We know that,

In an ideal transformer, the output power and input power are equal:

$$\begin{gathered} \Rightarrow {\mathrm{V}}_{\mathrm{s}}\times {\mathrm{i}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{p}}\times {\mathrm{i}}_{\mathrm{p}} \\ {\mathrm{i}}_{\mathrm{p}}={\mathrm{i}}_{\mathrm{s}}\times \frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}} \\ =4\times \frac{400}{200} \\ =4\times 2 \\ =8\mathrm{A} \end{gathered}$$

Therefore, the current in the primary is $8\mathrm{A}$.