Q131 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
The coefficient of mutual induction is $3$H and induced e.m.f. across secondary is $4$ kV. Current in primary is reduced from $7$A to $2$A. The time required for the change of current is
- Correct answerA$3.75 \times 10^{-3}$ s
- B$2.5 \times 10^{-3}$ s
- C$4.5 \times 10^{-3}$ s
- D$3.5 \times 10^{-3}$ s
Solution
The magnitude of induced e.m.f. in the secondary coil is given by
$|e| = M \left| \dfrac{\Delta I}{\Delta t} \right|$
Substituting the given values:
$4000 = 3 \times \left| \dfrac{2 - 7}{\Delta t} \right|$
$4000 = \dfrac{15}{\Delta t}$
$\Delta t = \dfrac{15}{4000} = 3.75 \times 10^{-3}$ s
Answer: $3.75 \times 10^{-3}$ s
$|e| = M \left| \dfrac{\Delta I}{\Delta t} \right|$
Substituting the given values:
$4000 = 3 \times \left| \dfrac{2 - 7}{\Delta t} \right|$
$4000 = \dfrac{15}{\Delta t}$
$\Delta t = \dfrac{15}{4000} = 3.75 \times 10^{-3}$ s
Answer: $3.75 \times 10^{-3}$ s
Hence, the correct answer is option: 'A'.