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PhysicsEMI/ACSelf and mutual inductorEasy2 minPYQ_2020

PhysicsEasysingle choice

The current in a coil of $L = 40 mH$ is to be increased uniformly from $1 A$ to $11 A$ in $4$ milliseconds. The induced e.m.f. will be

Answer:

Solution:

$e = L \frac{d I}{d t} = \frac{40 \times 10^{- 3} (11 - 1)}{4 \times 10^{- 3}} = 100 V$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Self and mutual inductor

⏱ 2mℹ️ Source: PYQ_2020

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