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Question

The primary coil of an idea step up transformer has $$100\ turns$$ and transformation ratio is also $$100$$. The input voltage and power are respectively $$220\ V$$ and $$1100\ W$$ calculate $$(2M)$$

1. Number of turns in secondary
2. Current in primary
3. Voltage across secondary
4. Current in secondary
5. Power in secondary


Solution

Given that,

Input voltage $${{V}_{input}}=220V$$

Power $${{P}_{input}}=1100\,W$$

Number of turns $${{N}_{P}}=100$$

$$r=100$$

(I). Number of turns in secondary

We know that,

  $$ r=\dfrac{{{N}_{s}}}{{{N}_{p}}} $$

 $$ {{N}_{_{S}}}=r\times {{N}_{p}} $$

 $$ {{N}_{s}}=100\times 100 $$

 $$ {{N}_{s}}=10000 $$

 

 (II).The current in primary

We know that

  $$ {{I}_{p}}=\dfrac{{{P}_{input}}}{{{V}_{p}}} $$

 $$ {{I}_{P}}=\dfrac{1100}{220} $$

 $$ {{I}_{p}}=5\,A $$


 (III). The Voltage across secondary

We know that

  $$ r=\dfrac{{{V}_{s}}}{{{V}_{p}}} $$

 $$ {{V}_{s}}=r\times {{V}_{p}} $$

 $$ {{V}_{s}}=100\times 220 $$

 $$ {{V}_{s}}=22000\,V $$

 

 (IV). The Current in secondary

We know that,

  $$ r=\dfrac{{{I}_{p}}}{{{I}_{S}}} $$

 $$ {{I}_{s}}=\dfrac{{{I}_{p}}}{r} $$

 $$ {{I}_{s}}=\dfrac{5}{100} $$

 $$ {{I}_{s}}=0.05\,A $$

 

(V). The power in secondary

We know that,

For ideal transformer

Power in secondary = power in primary

$${{P}_{s}}=1100\,W$$

This is the solution

 


Physics

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