Question

An ideal power transformer is used to step up an alternating $\text{EMF}$ of $220\text{V}$ to $4.4\text{kV}$ to transmit $6.6\text{kW}$ of power. If the primary coil has $1000$ turns, find the current in the secondary coil.

• A. $3\text{A}$
• B. $1.5\text{A}$
• C. $5\text{A}$
• D. $8.5\text{A}$

Answer 1

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

Power supplied at primary coil is given as

$i_p{E}_p=6.6\times {10}^3\text{W}$

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

For an ideal transformer we can use

$\frac{i_s}{i_p}=\frac{N_p}{N_s}=\frac{E_p}{E_s}$

Given that,

$E_p=220\text{V}$
$E_s=4.4\text{kV}=4400\text{V}$

$\Rightarrow \frac{i_s}{i_p}=\frac{220}{4400}$

$i_s=(\frac{1}{20}\times i_p)=(\frac{1}{20}\times 30)=1.5\text{A}$

Hence, option $\left(\text{B}\right)$ is correct.