A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power
Question:

A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of $2.2 \mathrm{~kW}$. If the current in the secondary coil is $10 \mathrm{~A}$, then the input voltage and current in the primary coil are :

  1. (1) $220 \mathrm{~V}$ and $20 \mathrm{~A}$

  2. (2) $440 \mathrm{~V}$ and $20 \mathrm{~A}$

  3. (3) $440 \mathrm{~V}$ and $5 \mathrm{~A}$

  4. (4) $220 \mathrm{~V}$ and $10 \mathrm{~A}$


Correct Option: , 3

Solution:

(3) Power output $\left(\mathrm{V}_{2} \mathrm{I}_{2}\right)=2.2 \mathrm{~kW}$

$\therefore \mathrm{V}_{2}=\frac{2.2 \mathrm{~kW}}{(10 \mathrm{~A})}=220 \mathrm{volts}$

$\therefore$ Input voltage for step-down transformer

$\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=2$

$\mathrm{V}_{\text {input }}=2 \times \mathrm{V}_{\text {output }}=2 \times 220$

$=440 \mathrm{~V}$

Also $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$

$\therefore \quad \mathrm{I}_{1}=\frac{1}{2} \times 10=5 \mathrm{~A}$

 

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