The flux linked with a circuit is given by
`phi=t^(3)+3t-5`
The graph between the indued e.m.f. (e) along the Y-axis and the time (t) along the X-axis will be

Options

A. a straight line through the origin
B. a straight line with a negative intercept on the e-axis
C. a straight line with a +ve intercept on the e-axis
D. a parabola not passing through the origin

Answer

D

Explanation

`phi= t^(3)+3t-5`
`:.` Induced e.m.f. `e= -(d phi)/(dt) = - d/(dt) (t^(3) + 3t -5)`
`e = -(3t^(2)+3) = -3(t^(2)+1)`
`:.` This is a second degree equation.
Hence the graph between e and t will be parabola.
When t = 0, e = -3 i.e. e != 0`
Thus the parabola will not pass through the origin.