Phasor Algebra

PHASOR ALGEBRA

To represent R, X_L, X_C in rectangular form:

Z_R = R + j0

Z_L = 0 + j X_L

Z_C = 0 - jX_C

To represent R, X_L, X_C in polar form :

Z_R =R ∠ 0°

Z_L = X_L ∠ 90°

Z_C = X_C ∠ -90°

Note:

1. In the rectangular form, if the reactive part (coefficient of j) is zero, the impedance is purely resistive.

2. In the rectangular form, if the active part is zero, the impedance is purely reactive. +j indicates inductive reactance -j indicates capacitive reactance.

3. In the polar form, if the angle is zero, the impedance is purely resistive.

4. In the polar form, if the angle is 90°, the impedance is purely reactive. +90° indicates inductive reactance and -90° indicates capacitive reactance.

For addition and subtraction, complex notation (rectangular) form is useful. For multiplication and division, polar form is necessary. So, the students should know conversion of rectangular to polar form and vice versa. For more details, the student can refer to the instructions of the Scientific Calculator.

Converting rectangular to polar form:

Example 1: Let Z = (10+ j15). Convert this into polar form. [Note: The author had used the scientific calculator CASIO fx 82B)

Solution: Steps:

1. Display 10

2. Press Shift key

3. Press R→P

4. Display 15.

5. Press = It shows 18.03 Ω

6. Press x →y It show the angle 56.31°

7. In the polar form, now Z is written as Z = 18.03 ∠56.31°

Example 2: Convert 20-j15 into polar form.

Solution: Steps:

1. Display 20

2. Press Shift key

3. Press R→P

4. Display -15.

5. Press = It shows 25 Ω

6. Press x → y It shows the angle -36.9°

7. In the polar form, now Z is written as Z = 25 ∠-36.9°

Note:

1. The student should not forget to put '-' in the reactive part.

2. The angle of Z always shows the phase angle ϕ. cos ϕ is called the power factor.

Converting polar to rectangular form

Example 3: Convert Z = 10 ∠60° into rectangular form.

Solution: Steps:

1. Display 10

2. Press Shift key

3. Press P →R

4. Display 60

5. Press = It show 5 (active part)

6. Press x → y It shows 8.66 (reactive part)

7. So, Z = 5 + j 8.66.

Example 4: Convert Z = 30 ∠45° into rectangular form.

Solution: Steps:

1. Display 30

2. Press Shift key

3. Press P→R

4. Display 45°

5. Press -/+

6. Press = It displays 21.21 (active part)

7. Press x→y It shows-21.22 (reactive part)

8. So, Z = 21.21-j21.21

Note:

1. The calculator must be set in 'deg' mode for conversions

2. In the polar form, if the angle is 90°, in the rectangular form active (real) part is 0.

3. In the polar form, if the angle is 45°, in the rectangular form, active and reactive parts are equal to each other.

4. Put j before reactive part.

Example 5 Z₁=10+j10 and Z₂ = 20-30 are in series. Find the Z_T in polar form.

Solution:

Z_T = Z₁+Z₂ = 10 + j10 + 20 - j30 = 30 - j20

Z_T = 36.1 ∠-33.7°

Note: The phase angle here is 33.7°. Since it is negative, the impedance is capacitive. So ϕ is leading in nature.

Example 6: The total impedance of a circuit in which Z₁ and Z₂ are in series is equal to (30+j40) Ω.

Z₁ = (20+j60) Ω. Find Z₂.

Solution: Steps:

Z_T = Z₁ + Z₂

Z₂ = Z_T - Z₁

= (30+j40) - (20+j60)

= 10-j20 {rectangular form}

= 22.36 ∠-63.43° {Polar form}

Example 7: In a given circuit I = 10∠60° Z = 20∠30. Find V.

Solution:

V = I. Z = (10 ∠ 60°) (20 ∠30°)

=(10 × 20) ∠ (60+30) = 200 ∠90°.

Example 8: In a circuit V = 200 volts I = 10∠30°. Find Z both in polar and rectangular forms.

Solution:

I = V/Z . It is division. So both V and I must be in polar form. Since angle of V is not given, we can take it as 0°, So,

V = 200 ∠ 0° and

I = 10 ∠ 30°

Z = V/I

= 200∠ 0°/10 ∠ 30°

= 200/10 ∠ (0-30)

=20∠-30° { Polar form}

=17.32-j10 { rectangular form}

Table 1:

Circuit response of single elements for DC:

Table 2: Voltage across each element for i = I_m sin ωt → I = |I| ∠20°

Where ϕ is the phase angle

Table 3: Current through each element when v = V_m sin ωt → V = V∠0°