Question

An ideal Transformer has 100 turns in the primary and 250 turns in the secondary. The peak value of the AC is $28\, V$. What is the nearest RMS secondary voltage?
A. $50\,V$
B. $70\,V$
C. $100\,V$
D. $40\,V$

Answer 1

The transformer equation is given by
$\dfrac{{{V_P}}}{{{V_S}}} = \dfrac{{{N_P}}}{{{N_S}}}$
Where the number of turns in the primary winding is denoted as ${N_P}$, the number of turns in the secondary winding is denoted as ${N_S}$, Voltage given to the primary winding is represented as ${V_P}$ and the voltage that we get at the secondary winding is represented as ${V_S}$.
The root means square voltage or RMS voltage represented as ${V_{rms}}$ is the square root of the time average of the square of the peak voltage.
The relation between peak voltage, ${V_0}$ and RMS voltage, ${V_{rms}}$ is given by the equation,
${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$

Complete step by step answer:
A transformer is a device that uses the process of electromagnetic induction to transfer electrical energy from one circuit to another. It can be used to increase the level of voltage or decrease the level of voltage between circuits.
A basic transformer consists of four primary parts.
1. input connection
2. output connection
3. core
4. the windings
A transformer has two windings, a primary winding, and a secondary winding. The number of turns in the primary winding is denoted as ${N_P}$. The number of turns in the secondary winding is denoted as ${N_S}$. Voltage given to the primary winding is represented as ${V_P}$ and voltage that we get at the secondary winding is represented as ${V_S}$. Then, the transformer equation is given by
$\dfrac{{{V_P}}}{{{V_S}}} = \dfrac{{{N_P}}}{{{N_S}}}$ …………………….(1)
Given,
${N_P} = 100$
${N_S} = 250$
${V_P}$=$28\, V$
Substitute the given values in equation (1)
$\Rightarrow \dfrac{{28\,V}}{{{V_S}}} = \dfrac{{100}}{{250}}$
$\Rightarrow{V_S} = \dfrac{{250}}{{100}} \times 28\,V$
On simplifications,
$\Rightarrow {V_S}= 70\,V$
In the question, we are asked to find the RMS value
The root means square voltage or RMS voltage represented as ${V_{rms}}$ is the square root of the time average of the square of the peak voltage.
The relation between peak voltage, ${V_0}$ and rms voltage, ${V_{rms}}$ is given by
${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$ ……………….(2)
Therefore, we can find the rms voltage at the secondary by using this equation. Then we get,
${V_{rms}} = \dfrac{{{V_S}}}{{\sqrt 2 }}$
Substituting $V_S$ value,
$\Rightarrow {V_{rms}}= \dfrac{{70\,V}}{{\sqrt 2 }}$
$\Rightarrow {V_{rms}}= 49.49\,V$
On approximation,
$\Rightarrow {V_{rms}} \approx 50\,V$

$\therefore$ The nearest RMS secondary voltage is $5V$. So, the correct answer is option A.

Note:
In the question the peak value of the voltage at the primary is given. So, when we substitute the values in the transformer equation. The value of the voltage that we get for the secondary will also be the peak value. Hence always remember to convert this peak value into rms value. If the value of the voltage at the primary is also given as rms voltage then by substituting that value in the transformer equation and solving we can directly get the rms voltage at the secondary.