Question

The output of a step down transformer is measured to be 48 V when connected to a 12 W bulb. What is the value of peak current?
\[
  {\text{A}}{\text{. }}\sqrt 2 {\text{ A}} \\
  {\text{B}}{\text{. }}\dfrac{1}{{2\sqrt 2 }}{\text{ A}} \\
  {\text{C}}{\text{. }}\dfrac{1}{4}{\text{ A}} \\
  {\text{D}}{\text{. }}\dfrac{1}{{\sqrt 2 }}{\text{ A}} \\
 \]

Answer 1

Hint: Here, we will proceed by finding out the current flowing through the secondary coils of the given step down transformer with the help of the given output voltage and given power. Then, we will use the relation between root mean square value of current and the peak current.
Formulas Used- ${\text{P}} = {\text{V}} \times {\text{I}}$ and ${{\text{I}}_{{\text{RMS}}}}$ = $\dfrac{{{{\text{I}}_0}}}{{\sqrt 2 }}$.

Complete Step-by-Step solution:
Given, Voltage output of a step down transformer ${{\text{V}}_{\text{s}}}$ = 48 V
Power of the bulb P = 12 W
As we know that the input voltage of any transformer (either step up or step down) is applied at the primary windings and the output voltage of that transformer is obtained across the secondary windings.
The secondary circuit of the given transformer is shown in the figure. Let the current passing through the secondary coils be I
Also we know that the relation between power P, voltage V and current I is given by
${\text{P}} = {\text{V}} \times {\text{I}}$
Using the above formula for the given secondary circuit, we can write
$
  {\text{P}} = {{\text{V}}_{\text{s}}} \times {\text{I}} \\
   \Rightarrow {\text{I}} = \dfrac{{\text{P}}}{{{{\text{V}}_{\text{s}}}}} \\
 $
By substituting ${{\text{V}}_{\text{s}}}$ = 48 V and P = 12 W in the above equation, we have
$ \Rightarrow {\text{I}} = \dfrac{{{\text{12}}}}{{{\text{48}}}} = \dfrac{1}{4}$ A
The above value of the current represents the current flowing in the given secondary circuit of the step down transformer.
Since, the current flowing through the secondary coil in the transformer is simply the root mean square value of the current i.e., ${{\text{I}}_{{\text{RMS}}}}$
i.e., ${{\text{I}}_{{\text{RMS}}}} = \dfrac{1}{4}$ A
Also, the relation between the peak current ${{\text{I}}_0}$ and root mean square value of the current is given by
${{\text{I}}_{{\text{RMS}}}}$ = $\dfrac{{{{\text{I}}_0}}}{{\sqrt 2 }}$
$ \Rightarrow {{\text{I}}_0} = \left( {\sqrt 2 } \right){{\text{I}}_{{\text{RMS}}}}$
By putting ${{\text{I}}_{{\text{RMS}}}} = \dfrac{1}{4}$ A in the above equation, we get
$
   \Rightarrow {{\text{I}}_0} = \left( {\sqrt 2 } \right)\left( {\dfrac{1}{4}} \right) \\
   \Rightarrow {{\text{I}}_0} = \dfrac{{\sqrt 2 }}{4} \\
 $
In order to rationalise the RHS of the above equation, we will multiply and divide the RHS by $\sqrt 2 $
\[
   \Rightarrow {{\text{I}}_0} = \dfrac{{\sqrt 2 }}{4} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} \\
   \Rightarrow {{\text{I}}_0} = \dfrac{2}{{4\sqrt 2 }} = \dfrac{1}{{2\sqrt 2 }}{\text{ A}} \\
 \]
Therefore, the value of the peak current is equal to \[\dfrac{1}{{2\sqrt 2 }}\] A

Hence, option B is correct.

Note- In a step up transformer, the output voltage obtained across the secondary windings is more than the input voltage applied across the primary windings whereas in case of a step down transformer, the output voltage obtained across the secondary windings is less than the input voltage applied across the primary windings.