Question

A transformer having an efficiency 80% is working on \(200\text{ V}\) and \(4\text{ kW}\) power supply. If the current in the secondary coil is \(5\text{ A}\), the voltage across the secondary coil and the current in the primary coil, respectively, are

• A. \(10\text{ A},600\text{ V}\)
• B. \(20\text{ A},640\text{ V}\)
• C. \(25\text{ A},680\text{ V}\)
• D. \(40\text{ A},700\text{ V}\)

Hint:

Primary current follows from input power and input voltage. The efficiency then gives the output power, which divided by the secondary current gives the secondary voltage.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Efficiency is the ratio of output power to input power. The primary current is found from input power divided by input voltage.

Step 2: Primary Current
\(I_p = \dfrac{P_{in}}{V_p} = \dfrac{4000}{200} = 20\text{ A}\)

Step 3: Output Power
\(P_{out} = 0.8 \times 4000 = 3200\text{ W}\)

Step 4: Secondary Voltage
\(V_s = \dfrac{P_{out}}{I_s} = \dfrac{3200}{5} = 640\text{ V}\)

Final Answer:
The primary current is 20 A and the secondary voltage is 640 V. \[ \boxed{\text{(B) }20\text{ A},640\text{ V}} \]