Question

An ideal transformer converts $220\, V$ a.c. to $3.3\, kV$ a.c. to transmit a power of $4.4\, kW$. If primary coil has $600$ turns, then alternating current in secondary coil is

• A. $\frac{1}{3} A$
• B. $\frac{4}{3} A$
• C. $\frac{5}{3} A$
• D. $\frac{7}{3} A$

Answer 1

Answer: (B)

$\frac{4}{3} A$

Answer 2

Given, input voltage $\left(V_{P}\right)=220 V$
Output voltage $\left(V_{S}\right)=3.3 \times 10^{3} V$
Power$(P)=4.4 kW =4.4 \times 10^{3} W$
Number of turns in primary coil $=600$
We know that,
$P=V_{P} \times I_{P} \Rightarrow I_{P}=\frac{P}{V_{P}}$ or $I_{P}=\frac{4.4 \times 10^{3}}{220}=20 A$
For an ideal transformer,
$\frac{V_{S}}{V_{P}}=\frac{I_{P}}{I_{S}}$
$I_{S}=I_{P} \times \frac{V_{P}}{V_{S}}$
$I_{S}=\frac{20 \times 220}{3.3 \times 10^{3}}$
Or $\,\,\,\,\,\,\,\, I_{S}=\frac{44}{33}=\frac{4}{3} A$