Question

An ideal transformer , rated at $6.6\text{kW}$, is used to step up an alternating voltage having peak value of $220\text{V}$ to $4.4\text{kV}$. If the primary coil has $1,000$ turns, find the current rating of the secondary coil.

• A. $3.5\text{A}$
• B. $1.5\text{A}$
• C. $5.5\text{A}$
• D. $8.5\text{A}$

Answer 1

The correct option is B $1.5\text{A}$

Power supplied at primary coil is given by,

$P_{\text{input}}=E_P{i}_P=6.6\times {10}^3\text{W}$

$\Rightarrow i_P=\frac{6.6\times {10}^3}{220}=30\text{A}$

Now, for an ideal transformer,

$\frac{i_P}{i_S}=\frac{N_S}{N_P}=\frac{E_S}{E_P}$

$\Rightarrow i_S=\frac{E_P}{E_S}\times i_P=\frac{220}{4400}\times 30=1.5\text{A}$

Hence, option $\left(B\right)$ is the correct answer.