Question

What is the value of peak voltage of a centre tapped transformer rated 12-0-12 V?
A. $12V$
B. $\dfrac{{12}}{{\sqrt 2 }}V$
C. $12\sqrt 2 V$
D. $24V$
E. $24\sqrt 2 V$

Answer 1

Hint:The highest value of any voltage waveform is called the peak voltage.
The peak voltage for transformer in terms of root mean square value of voltage is expressed as-
${V_p} = {V_{rms}}\sqrt 2 $

Complete step-by-step solution:
The transformer is an electrical device which transmits electrical energy from one circuit to another circuit which works on the basis of electromagnetic induction and mutual induction.
The working principle of a transformer is based on the mutual induction in two or more coils which transfers electrical energy between circuits.
Mutual induction is the generation of current from one coil to another by changing the current of one coil.
There are two types of transformer: Step – up transformer and Step – down transformer
The transformer which increases voltage between primary to secondary windings is called Step – up transformer and the transformer which decreases voltage between primary to secondary windings is called Step – down transformer.
Peak voltage is the highest value of any voltage waveform. The peak voltage is obtained by increasing impedance and switching speed. Peak voltage is measured from the horizontal axis, from the top of the waveform which is called the crest of the waveform.
According to question it is given that,
R.M.S value of voltage is $12V$
Let the rms value of voltage be ${V_{rms}}$ and peak voltage be ${V_p}$.
The peak voltage can be calculated by-
${V_p} = {V_{rms}}\sqrt 2 $
Putting the value of rms value of voltage in above formula-
${V_p} = 12\sqrt 2 V$
Hence, the correct option is (C).

Note:- Peak voltage can be calculated by peak – to – peak voltage, rms voltage and average voltage.
->Calculating peak voltage from peak to peak voltage-
${V_p} = {V_{ptp}} \times 0.5$
where, ${V_{ptp}}$ is peak to peak voltage
->Calculating peak voltage from RMS voltage
${V_p} = {V_{rms}} \times \sqrt 2 $
->Calculating peak voltage from average voltage
${V_p} = {V_{average}} \times \dfrac{\pi }{2}$