A transformer having efficiency of 75 \% is working on 220 V and...

A transformer having efficiency of

75 \%

is working on 220 V and 4.4 kW power supply. If the current in the secondary coil is 5 A . What will be the voltage across secondary coil and the current in primary coil?

Options

A $V_S=220 \mathrm{~V}, i_P=20 \mathrm{~A}$
B $V_S=660 \mathrm{~V}, i_P=15 \mathrm{~A}$
C $V_S=660 \mathrm{~V}, i_P=20 \mathrm{~A}$
D $V_S=220 \mathrm{~V}, i_P=15 \mathrm{~A}$

Correct answer

C. $V_S=660 \mathrm{~V}, i_P=20 \mathrm{~A}$

Step-by-step solution

Given,

P_{\text{in }}=4.4 \mathrm{~kW}, V_p=220 \mathrm{~V}, i_s=5 \mathrm{~A}, V_s=

? We know that,

\quad i_P=\frac{P_{\text{in }}}{V_P}=\frac{4.4 \times 10^3}{220}=20 \mathrm{~A}\therefore

Efficiency,

\eta=\frac{\text{ Output power }}{\text{ Input power }}

Percentage efficiency

\eta=\frac{P_{\text{out }}}{P_{\text{in }}} \times 100

\begin{aligned} 75 \% & =100 \times \frac{P_{\text{out }}}{4.4 \times 10^3} \\ \therefore \quad P_{\text{out }} & =\frac{75 \times 4.4 \times 10^3}{100} \\ & =3300 \mathrm{~W} \\ \therefore \quad P_{\text{out }} & =V_S \cdot i_S \\ \Rightarrow \quad V_S & =\frac{3300}{5}=660 \mathrm{~V}\end{aligned}

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