Question

Convert the following instantaneous current to phasors, using cos(wt) as the reference. Give your answers in both rectangular and polar form.

Short answer

(a) The instantaneous current in polar form is 500∠-30^o and in rectangular form is (612.j353.55)

(b) The instantaneous current in polar form 4∠-60^o is and in rectangular form is (2-j3.464)

(c) The instantaneous current in polar form 9.832∠-39^o is and in rectangular form is (7.657-j6.192).

Step-by-step solution

Write the given data of the question.

The instantaneous value of the current in part-a is given as i (t) =500√2 cos (wt -30^o)

The instantaneous value of the current in part-a is given as 4sin (wt +30^o),

The instantaneous value of the current in part-a is given as i (t) =5cos(wt-15^o) + 4√2 sin (wt -30^o) ,

Write the concept for polar form and rectangular form.

In case of the rectangular form a complex number can be represented into the graph by real (horizontal axis) and imaginary axis (vertical axis). For example z=3+j8

here, 3 represent magnitude of the real axis points and represents magnitude of the vertical axis points.

The polar form is represented by magnitude and angle. The polar form is also known by the name of phasor form.

If x (t)=Acoswt is taken as the reference for the phasor diagram.

Then the phasor diagram for is, y(t)=Acos (wt -Θ) is

The polar form of y (t) is,

y (t) =A∠-Θ

Here-Θ means that y (t) lags x (t) byΘ angle

The rectangular form of y (t) is,

Calculate the rectangular and polar form in part-a.

(a)

Let us assume as the reference;

Then the phasor diagram for is i (t) =500√2 cos (wt -30^o) is,

Consider the Polar form of i (t) is,

i (t) =500√2 ∠-30^o

Here-30^o means that the current i (t) lags by 30^o .

The rectangular form of i (t) is calculated as,

Calculate the rectangular and polar form in part-b.

(b)

Let us assume as the reference,

sin θ = cos (90^o- θ )

Using the above concept;

Then the phasor diagram for 4 sin (wt +30^o) is,

Write the polar form.

i (t) = 4∠ -60^o

Here -60^o means that the current i (t) lags by 60^o

Write the rectangular form as,

Therefore the polar form is 4∠ -60^o and the rectangular form is (2-j3.464) .

Calculate the rectangular and polar form in part-c.

(c)

Let us assume coswt as the reference

As

sin θ= cos (90^o-θ)

Using the above concept

Given equation of instantaneous current is,

The Polar form of i (t) is,

Overall polar form after resolving,

Here -15^o means that the current i (t) lags by 15^o and -60^o means that the current i (t) lags by 60^o

Write the rectangular form as,