The emf phasor in the figure(Figure 1) is shown at t = 12ms . Assume E0 = 18V . Part A What is the angular?

The emf phasor in the figure(Figure 1) is shown at t = 12 ms . Assume E0 = 18V .

Answer 1

A

From the diagram you can tell the phasor has rotated 120° (because, by convention, the start position is taken as pointing to the right, along the +x axis, and the direction of rotation is taken as anticlockwise).

120° is ⅓ of a rotation (because 120°/360° = ⅓ ).

So the phasor has rotated ⅓ of a rotation in 12ms.
One full rotation takes 12/⅓ = 36ms. = 0.036s. This is the period,

ω = 2π/T = 2π/0.036 = 174.5rad/s.
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A quicker method is ω = (angle in radians)/(time).
ω = θ/t

Since θ = 120° = 2π/3 radians, ω = (2π/3)/0.012 = 174.5rad/s.
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B

emf = E₀sin(θ) = 18sin((120°) = 15.6v

If you prefer, you can use
emf = E₀sin(ωt) = 18sin(174.5x 0.012) = 18sin(2.094)
Remember if you are doing it this way, 2.094 is in radians, so switch your calculator to radians mode
emf = 18sin(2.094) = 15.6V

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