The current through a coil of self-inductance L = 2mH is given by i = t²e^–t at time t. How long it will take to make the emf zero?
Added by Sotor S.
Step 1
Step 1: The equation for the EMF in the circuit is given by EMF = -L(di/dt), where L is the self-inductance (2mH = 2x10^-3 H) and di/dt is the rate of change of current with respect to time. Show more…
Show all steps
Your feedback will help us improve your experience
Kamlesh Goyal and 62 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The current (in Ampere) in an inductor is given by I = 5t + 16t , where is in seconds. The self-induced emf in it is 10mV. The self-inductance will be ?
Krishna J.
At the instant when the current in an inductor is increasing at a rate of 0.0640 $\mathrm{A} / \mathrm{s}$ , the magnitude of the self-induced emf is 0.0160 $\mathrm{V} .$ What is the inductance of the inductor?
For a coil having $L=2 \mathrm{mH}$, current flow through it is $I$ $=t^{2} e^{-t}$ then, the time at which emf become zero (a) $2 \mathrm{sec}$ (b) $1 \mathrm{sec}$ (c) $4 \mathrm{sec}$ (d) $3 \mathrm{sec}$
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD