A 14.8-mH inductor carries a current i = Imax sin ωt, with Imax = 5.40 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a function of time? (Express your answer in terms of t where e m f is in volts and t is in seconds. Do not include units in your expression.)
this is great question, get answer with step by step process
www.careerorbits.com/jee-main/mock-test-series-online-cbt
To find the self-induced emf as a function of time, we can use Faraday's Law of electromagnetic induction, which states that the emf induced in a coil is equal to the rate of change of magnetic flux through the coil.
The self-induced emf can be calculated using the formula:
emf = -L(dI/dt)
where emf is the self-induced emf, L is the inductance of the coil, and (dI/dt) is the time rate of change of current flowing through the coil.
In this case, the inductor has an inductance L = 14.8 mH = 14.8 * 10^(-3) H.
Given that the current i = Imax sin(ωt), where Imax = 5.40 A and f = ω/2π = 60.0 Hz, we can find the time rate of change of current by taking the derivative of the current function with respect to time:
(dI/dt) = Imax ω cos(ωt)
Substituting the given values into the equation, we have:
(dI/dt) = (5.40 A) * (2π * 60.0 Hz) * cos(2π * 60.0 Hz * t)
Simplifying this expression, we get:
(dI/dt) = (5.40 A) * (120π) * cos(120πt)
Finally, substituting the values of L and (dI/dt) into the formula for the self-induced emf, we have:
emf = - (14.8 * 10^(-3) H) * (5.40 A) * (120π) * cos(120πt)
Simplifying this expression, the self-induced emf as a function of time is:
emf = - 95.299 π cos(120πt)
Therefore, the self-induced emf as a function of time is - 95.299 π cos(120πt) volts.