
A step down transformer, transforms a supply line voltage of 2200 V into 220 V. The primary coil has 5000 turns. The efficiency and the power transmitted by the transformer are 90% and 8 KW respectively. Then the power supplied is
$\left( A \right)$ 9.89 KW
$\left( B \right)$ 8.89 KW
$\left( C \right)$ 88.9 KW
$\left( D \right)$ 889 KW
Answer
123k+ views
- Hint: In this question use the property that efficiency of a transformer is the ratio of power received at the load end to the power supplied by the source end so use this concept to reach the solution of the question.
Complete step-by-step solution -
Given data:
Supply voltage = 2200 V
Load voltage = 220 V
Number of turns in primary coil = 500
Efficiency of transformer = 90% = (90/100)
Power transmitted i.e. power received by the load end = 8 KW
Now as we know that efficiency ($\eta $) of a transformer is the ratio of power received (${P_r}$) at the load end to the power supplied (${P_s}$) by the source end.
$ \Rightarrow \eta = \dfrac{{{P_r}}}{{{P_s}}}$
Now substitute the values we have,
$ \Rightarrow \dfrac{{90}}{{100}} = \dfrac{{8{\text{ KW}}}}{{{P_s}}}$
$ \Rightarrow {P_s} = \dfrac{{8{\text{ KW}}}}{{\dfrac{{90}}{{100}}}} = \dfrac{{800}}{{90}}{\text{ = 8}}{\text{.89 KW}}$
So the power supplied by the transformer is 8.89 KW.
So this is the required answer.
Hence option (B) is the correct answer.
Note – An interesting fact is that for an ideal transformer the power supplied by the transformer is equal to the power received as the efficiency of the ideal transformer is 100%, if not than it is not an ideal transformer it is a practical transformer and the loss in the power while transferring the power from source to load (i.e. the difference of power) is called as core loss and copper loss of the transformer.
Complete step-by-step solution -
Given data:
Supply voltage = 2200 V
Load voltage = 220 V
Number of turns in primary coil = 500
Efficiency of transformer = 90% = (90/100)
Power transmitted i.e. power received by the load end = 8 KW
Now as we know that efficiency ($\eta $) of a transformer is the ratio of power received (${P_r}$) at the load end to the power supplied (${P_s}$) by the source end.
$ \Rightarrow \eta = \dfrac{{{P_r}}}{{{P_s}}}$
Now substitute the values we have,
$ \Rightarrow \dfrac{{90}}{{100}} = \dfrac{{8{\text{ KW}}}}{{{P_s}}}$
$ \Rightarrow {P_s} = \dfrac{{8{\text{ KW}}}}{{\dfrac{{90}}{{100}}}} = \dfrac{{800}}{{90}}{\text{ = 8}}{\text{.89 KW}}$
So the power supplied by the transformer is 8.89 KW.
So this is the required answer.
Hence option (B) is the correct answer.
Note – An interesting fact is that for an ideal transformer the power supplied by the transformer is equal to the power received as the efficiency of the ideal transformer is 100%, if not than it is not an ideal transformer it is a practical transformer and the loss in the power while transferring the power from source to load (i.e. the difference of power) is called as core loss and copper loss of the transformer.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Main Chemistry Exam Pattern 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main
