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 the secondary coil is

• A. \(\frac{4}{3}\) A
• B. \(\frac{3}{4}\) A
• C. \(\frac{1}{3}\) A
• D. \(\frac{2}{3}\) A

Hint:

For an ideal transformer, output power equals the transmitted power, so I = P/V on the secondary.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An ideal transformer has no loss, so input power equals output power. The power delivered by the secondary is 4.4 kW at 3.3 kV.

Step 2: Find the secondary current:
\(I_s=\dfrac{P}{V_s}=\dfrac{4400}{3300}=\dfrac43\) A. Option A.

Step 3: Check with the turns ratio:
The primary current is \(I_p=\dfrac{4400}{220}=20\) A. The turns ratio is \(\dfrac{N_s}{N_p}=\dfrac{3300}{220}=15\), so \(I_s=\dfrac{I_p}{15}=\dfrac43\) A. This agrees.

Step 4: Why the other options are wrong.
3/4, 1/3 and 2/3 A would give powers of 2.5 kW, 1.1 kW and 2.2 kW, none equal to 4.4 kW.

Final Answer:
The secondary current is 4/3 A. \[ \boxed{\text{(A) }\dfrac43\ \text{A}} \]