
A power transmission line feeds input power at $2300{\text{V}}$ to a step down transformer with its primary windings having ${\text{4000}}$ turns. What should be the number of turns in the secondary in order to get output power at $230{\text{V}}$?
Answer
131.4k+ views
Hint: The voltage at the either side of a transformer is proportional to the number of turns at that side. Therefore, the ratio of the voltages at the primary and the secondary side of a transformer is equal to the ratio of the number of turns of the primary and the secondary windings. Using this relation, we can substitute the given values and can determine the required value of the number of turns in the secondary.
Formula used: The formula used to solve this question is given by
$\dfrac{{{V_1}}}{{{N_1}}} = \dfrac{{{V_2}}}{{{N_2}}}$, here ${V_1}$ and ${N_1}$ are the voltage and the number of turns at the primary of a transformer, and ${V_2}$ and ${N_2}$ are the corresponding values at the secondary.
Complete step-by-step solution:
Let the number of turns in the primary and the secondary windings of the given transformer be ${N_1}$ and ${N_2}$ respectively. Also, let ${V_1}$ and ${V_2}$ be the respective values of the voltages at the primary and the secondary of the transformer.
According to the question, we have
${V_1} = 2300{\text{V}}$................(1)
${V_2} = 230{\text{V}}$ ………...(2)
Also,
${N_1} = 4000$ …………………..(3)
We know that for a transformer we can write
$\dfrac{{{V_1}}}{{{N_1}}} = \dfrac{{{V_2}}}{{{N_2}}}$
Substituting (1), (2) and (3) in the above equation we get
$\dfrac{{2300}}{{4000}} = \dfrac{{230}}{{{N_2}}}$
$ \Rightarrow {N_2} = 230 \times \dfrac{{4000}}{{2300}}$
On solving we get
${N_2} = 400$
Hence, for the given value of the output voltage, the number of turns in the secondary of the given transformer is equal to $400$.
Note: A step down transformer is used to decrease the high voltage applied at the primary side to the low voltage. For this purpose it has less number of turns at the secondary side than at the primary side. Since the given transformer is stepped down, confirm that the number of turns at the secondary that you have got must be less than the number of turns at the primary.
Formula used: The formula used to solve this question is given by
$\dfrac{{{V_1}}}{{{N_1}}} = \dfrac{{{V_2}}}{{{N_2}}}$, here ${V_1}$ and ${N_1}$ are the voltage and the number of turns at the primary of a transformer, and ${V_2}$ and ${N_2}$ are the corresponding values at the secondary.
Complete step-by-step solution:
Let the number of turns in the primary and the secondary windings of the given transformer be ${N_1}$ and ${N_2}$ respectively. Also, let ${V_1}$ and ${V_2}$ be the respective values of the voltages at the primary and the secondary of the transformer.
According to the question, we have
${V_1} = 2300{\text{V}}$................(1)
${V_2} = 230{\text{V}}$ ………...(2)
Also,
${N_1} = 4000$ …………………..(3)
We know that for a transformer we can write
$\dfrac{{{V_1}}}{{{N_1}}} = \dfrac{{{V_2}}}{{{N_2}}}$
Substituting (1), (2) and (3) in the above equation we get
$\dfrac{{2300}}{{4000}} = \dfrac{{230}}{{{N_2}}}$
$ \Rightarrow {N_2} = 230 \times \dfrac{{4000}}{{2300}}$
On solving we get
${N_2} = 400$
Hence, for the given value of the output voltage, the number of turns in the secondary of the given transformer is equal to $400$.
Note: A step down transformer is used to decrease the high voltage applied at the primary side to the low voltage. For this purpose it has less number of turns at the secondary side than at the primary side. Since the given transformer is stepped down, confirm that the number of turns at the secondary that you have got must be less than the number of turns at the primary.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Clemmenson and Wolff Kishner Reductions for JEE

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

In Bohrs model of the hydrogen atom the radius of the class 12 physics JEE_Main

JEE Main 2025 Session 2 Registration Open – Apply Now! Form Link, Last Date and Fees

Other Pages
JEE Advanced 2024 Syllabus Weightage

CBSE Date Sheet 2025 Class 12 - Download Timetable PDF for FREE Now

JEE Main 2025 - Session 2 Registration Open | Exam Dates, Answer Key, PDF

CBSE Class 10 Hindi Sample Papers with Solutions 2024-25 FREE PDF

CBSE Board Exam Date Sheet Class 10 2025 (OUT): Download Exam Dates PDF

CBSE Class 10 Hindi Course-B Syllabus 2024-25 - Revised PDF Download
