Final answer:
To draw the phasor for the given EMF at t=60ms, first calculate the angular frequency, then the phase angle at this time, which shows the phasor is horizontal with a magnitude of 170V.
Step-by-step explanation:
The student has asked to draw the phasor for the EMF e=(170V)cos((2π×60Hz)t) at t=60ms. To draw the phasor, we need to calculate the angle the phasor makes with the horizontal axis at the given time. The angular frequency (ω) is 2π×60Hz. Therefore, at t=60ms or 0.060s, the phase angle θ is ωt = (2π×60Hz) × 0.060s = 36π rad or 6 rotations. Since 6 rotations bring the phasor back to the same position, its angle with respect to the horizontal axis is 0 degrees.
The magnitude of the phasor is the amplitude of the EMF, which is 170V. So, we draw a horizontal phasor with a length that represents 170V, pointing to the right which corresponds to an angle of 0 degrees at t=60ms.