MHT CET · Physics · Alternating Current

Alternating current of peak value $(\frac{2}{π})$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

A 100 V
B 200 V
C 300 V
D 400 V

Verified Solution

Answer & Solution

Correct Answer

(B) 200 V

Step-by-step Solution

Detailed explanation

Peak value of current

I_{0} = I_{r m s} \times \sqrt{2} = (\frac{2}{π})

Amp
Co-efficient of mutual inductance is M = 1 henry
Induced emf in secondary is given by

e_{2} = M \cdot \frac{d i}{d t}

where

i = i_{0} . \sin (ω t + ϕ)

Here

ω= 2 π n = 100 π .

∴ e_{2} = 1 \times \frac{d}{d t} (i_{0} . \sin ω t)

i_{0} . ω . \cos (ω t) = \frac{2}{π} \times 2 π \times 50 . \cos (100 π t)

For,

t = 0

, we have

e_{2} = 4 \times 50 = 200 V

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Alternating current of peak value 2π ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

Step-by-step Solution

Alternating current of peak value $(\frac{2}{π})$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)