A transformer consisting of 300 turns in the primary and 150 turns in secondary 

gives output power of 2.2 kW. If the current in the secondary coil is 10A, then the 

input voltage and current in the primary coil are:

  • Option 1)

    220 V and 20 A 

  • Option 2)

    440 V and 20 A

  • Option 3)

    440 V and 5 A

  • Option 4)

     220 V and 10 A

 

Answers (1)

 

Ideal transformer -

There is no loss of power.

- wherein

P_{out}= P_{in}

V_{s}\, {i}'_{s}= V_{in}i_{in}

 

 

Ideal transformer -

\frac{\varepsilon _{s}}{\varepsilon _{P}}= \frac{N _{s}}{N _{P}}= \frac{V_{s}}{V _{P}}= \frac{{i}' _{P}}{{i}'_{s}}
 

- wherein

\varepsilon _{s}\rightarrow induced e.m.s.in secondary

\varepsilon _{P}\rightarrow induced e.m.s.in primary

 

 

P_{out}=V_oI_o

P_{out}=2.2kW

=> 2.2kW=V_o\times 10

=> V_o=\frac{2.2\times 10^{3}}{10}=220V

Now,

\frac{n_I}{n_{II}}=\frac{V_I}{V_{II}}=\frac{I_{II}}{I_I}

So, \frac{300}{150}=\frac{V_I}{220}

=> V_I=440V

Similarly,

\frac{300}{150}=\frac{I_{II}}{I_{I}}

I_{I}=\frac{I_{II}}{2}=\frac{10}{2}=5A


Option 1)

220 V and 20 A 

Option 2)

440 V and 20 A

Option 3)

440 V and 5 A

Option 4)

 220 V and 10 A

Preparation Products

Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Test Series JEE Main July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 17999/- ₹ 11999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 19999/-
Buy Now
Exams
Articles
Questions