Question

A transformer of efficiency $90$% has turns ratio $1:10$. If the voltage across the primary is $220V$ and current in the primary is $0.5A$, then the current in secondary is

• A. $5.5A$
• B. $5A$
• C. $4A$
• D. $4.5A$

Answer 1

The correct option is D $4.5A$

Given : $\frac{N_s}{N_p}=\frac{1}{10}$ $E_p=220V$ $I_p=0.5A$

Efficiency of transformer $\eta =\frac{P_{out}}{P_{in}}=\frac{E_s{I}_s}{E_p{I}_p}$

$\therefore$ $0.9=\frac{E_s{I}_s}{E_p{I}_p}$ ......(1)

Using $\frac{E_s}{E_p}=\frac{N_s}{N_p}=\frac{1}{10}$

So, $0.9=\frac{1\times I_s}{10\times 0.5}$

$⟹I_s=4.5A$