Question

A transformer has an efficiency of $80$ percentage. It is connected to a power input of $5KW$ at $200V$ If the secondary voltage is $250V$, the primary and secondary current are respectively
(A) $25A,20A$
(B) $20A,16A$
(C) $25A,16A$
(D) $40A,25A$
(E) $40A,16A$

Answer 1

Hint: In order to solve this question, we will first calculate the primary current using input voltage and power relation, and then using the efficiency of the transformer we will solve for secondary current using output power and voltage relation.

Formula Used:
Primary current is given by the ratio of input power and input voltage,
${I_P} = \dfrac{{{P_{input}}}}{{{V_{input}}}}$
Efficiency is equal to ratio of Output power to primary power as,
$\eta = \dfrac{{{P_{output}}}}{{{P_{input}}}}$

Complete step by step solution:
We have given that the input power as ${P_{input}} = 5KW = 5000W$ and input voltage is ${V_{input}} = 200V$
Using the formula for primary current as,
${I_P} = \dfrac{{{P_{input}}}}{{{V_{input}}}}$
We get,
${I_P} = \dfrac{{5000}}{{200}} \\
\Rightarrow {I_P} = 25A \\ $
Hence, the primary current is $25A$.

Now, Using Efficiency formula we have given that $\eta = 80$ percentage and input power as ${P_{input}} = 5000$ then output power is calculated using $\eta = \dfrac{{{P_{output}}}}{{{P_{input}}}}$ we get,
$\dfrac{{80}}{{100}} = \dfrac{{{P_{output}}}}{{5000}} \\
\Rightarrow {P_{output}} = 4000W \\ $
Now using formula, ${I_S} = \dfrac{{{P_{output}}}}{{{V_{output}}}}$ we get the secondary current as (C) $25A,16A$
${I_S} = \dfrac{{4000}}{{25}} \\
\Rightarrow {I_s} = 16A \\ $
Hence, the secondary current is ${I_s} = 16A$

Hence, the correct option is (C) $25A,16A$.

Note: It should be remembered that, always convert the units of Power from Kilowatts to Watts and its related as $1KW = 1000W$ and efficiency must be converted from percentage value into its numerical value.