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.

Open in App
Solution

## Step 1: Given informationIn 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 findWe have to find the current in the primary winding.Step 3: CalculationWe know the relation,$\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}}=\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}}$or,${\mathrm{V}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{p}}×\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}}$Here,Vp is the primary voltage.Vs is the secondary voltage.Ns is the secondary turns.Np is the primary turns.By substituting the given values, we get${\mathrm{V}}_{\mathrm{s}}=200×\frac{1000}{500}\phantom{\rule{0ex}{0ex}}=400\mathrm{V}$The current in the secondary winding will be:${\mathrm{i}}_{\mathrm{s}}=\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{R}}_{\mathrm{s}}}$Here, Rs is the load resistance.By substituting the values, $=\frac{400}{100}\phantom{\rule{0ex}{0ex}}=4\mathrm{A}$We know that,In an ideal transformer, the output power and input power are equal:$⇒{\mathrm{V}}_{\mathrm{s}}×{\mathrm{i}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{p}}×{\mathrm{i}}_{\mathrm{p}}\phantom{\rule{0ex}{0ex}}{\mathrm{i}}_{\mathrm{p}}={\mathrm{i}}_{\mathrm{s}}×\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}}\phantom{\rule{0ex}{0ex}}=4×\frac{400}{200}\phantom{\rule{0ex}{0ex}}=4×2\phantom{\rule{0ex}{0ex}}=8\mathrm{A}$Therefore, the current in the primary is $8\mathrm{A}$.

Suggest Corrections
0
Explore more