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: Use current ratio:
For an ideal transformer, $\dfrac{I_s}{I_p}=\dfrac{V_p}{V_s}=\dfrac{220}{3300}=\dfrac1{15}$.

Step 2: Primary current:
$I_p=\dfrac{P}{V_p}=\dfrac{4400}{220}=20$ A.

Step 3: Secondary current:
$I_s=\dfrac{20}{15}=\dfrac43$ A. Option A. The 600 primary turns are not needed for this result.

Final Answer:
Power 4.4 kW at 3.3 kV gives I = 4/3 A. \[ \boxed{\text{(A) }\dfrac43\ \text{A}} \]