The current drawn by the primary of a transformer, which steps down to $20\,V$ to operate a device of resistance $20\Omega $ is: (Assume the efficiency of the transformer to $80\% $ )
A. $0.125\,A$
B. $0.225\,A$
C. $0.325\,A$
D. $0.425\,A$

Hint: To solve this question, we must have a knowledge of transformers and here we simply used the concept of efficiency of transformers in which we simply put the values in the formula and solve accordingly and hence we get the required solution. The ratio of output power to the input power is called the efficiency of the transformer.

Formula used:
$\eta = \dfrac{{{\text{power output}}}}{{{\text{power input}}}}$

Complete step by step answer:
According to the question, the primary voltage of the transformer is ${V_p} = 200V$. Secondary voltage of the transformer is ${V_s} = 20V$, resistance is $R = 20\Omega $ and efficiency of the transformer is $80\% = 0.8$. And we know that the efficiency of a transformer is given by,
$\eta = \dfrac{{{\text{power output}}}}{{{\text{power input}}}} \\
\Rightarrow \eta = \dfrac{{{V_s} \times {i_s}}}{{{V_p} \times {i_p}}}$
And we know that,
\[{i_s} = \dfrac{{{V_s}}}{R} \\
\Rightarrow {i_s} = \dfrac{{20}}{{20}} \\
\Rightarrow {i_s} = 1\,A\]
Now simply substituting all the values in the above formula and solving for the current,
$\eta = \dfrac{{{V_s} \times {i_s}}}{{{V_p} \times {i_p}}} \\
\Rightarrow 0.8 = \dfrac{{20\,V \times 1\,A}}{{200\,V \times {i_p}}} \\
\Rightarrow {i_p} = \dfrac{{20\,V \times 1\,A}}{{200\,V \times 0.8}} \\ $
On further solving we get,
$ \therefore {i_p} = 0.125\,A$

So, the current drawn by the primary of a transformer is $0.125\,A$.

Note: Note that all the units should be in standard form and if not, first you have to convert and then proceed and remember the formula. A step-down transformer transforms a high primary voltage to a low secondary voltage. In a step-down transformer there are more turns in a primary winding than that of secondary.