Question

A 50 Hz AC current of crest value 1 A flows through the primary of a transformer. If the mutual inductance between the primary and secondary is 0.5 H, the crest voltage induced in the secondary is______.
Fill in the blank with the correct answer from the options given below.

• A. 75 V
• B. 150 V
• C. 100 V
• D. 200 V

Answer 1

The Correct Option is B

Let's analyze the induced voltage in the secondary of the transformer.

1. Induced Voltage (ε_s):

The induced voltage in the secondary is given by:

ε_s = M(dI_p/dt)

Where:

• M is the mutual inductance
• I_p is the current in the primary
• t is time

2. Primary Current (I_p):

The primary current is given by:

I_p = I₀sin(ωt)

Where:

• I₀ is the crest value of the current (1 A)
• ω is the angular frequency (ω = 2πf)
• f is the frequency (50 Hz)

Therefore:

I_p = 1 * sin(2π * 50 * t)

I_p = sin(100πt)

3. Rate of Change of Primary Current (dI_p/dt):

dI_p/dt = d(sin(100πt))/dt

dI_p/dt = 100πcos(100πt)

4. Induced Voltage (ε_s):

ε_s = M(dI_p/dt)

ε_s = 0.5 H * 100πcos(100πt)

ε_s = 50πcos(100πt)

5. Crest Voltage (ε₀):

The crest voltage is the maximum value of ε_s, which occurs when cos(100πt) = 1:

ε₀ = 50π V

ε₀ = 50 * 3.14 V

ε₀ = 157 V

Therefore, the crest voltage induced in the secondary is approximately 157 V.

The closest answer is 150V.

The correct answer is:

Option 2: 150 V