Introduction to Electric Circuits

In polar form

A = V/(5 2 --I-- 6 2) /tan -~ (6/5) - 7.81/_50.2~ V

and

B = ~/(72 + 102)/_ - tan -~ (10//7) = 12.21/_-55~ A

A x B - AB/--(ChA + ~B) -- 7.81 X 12.21/(50.2 -- 55)

= 95.33/_-4.8 ~

Example 4.17

4.4 Complex notation 89

Divide A : (5 + j6) by B - (7 -jl0).

Solution

Using rectangular form we have A/B = (5 + j6)/(7 - jl0). To proceed from

here we have to 'rationalize' the denominator, i.e. remove the j. This is done by

multiplying the numerator and the denominator by the conjugate of the

denominator. The conjugate of a complex number (a + jb) is (a - jb), obtained

simply by changing the sign of the j term. When multiplying a complex number

by its conjugate the j disappears. Thus

(7 -j10)(7 + jl0) = (7 • 7) + (7 • jl0) -j(10 • 7) -j:(10 • 10)

= 49 + j70 -j70 - (-1)(100) = 49 + 100

= 149

Hence

A/B = [(5 + j6)(7 + j10)]/[(7 -j10)(7 + jl0)]
= (35 + j50 + j42 + j260)/149
=(-25 + j92)//149
= (-0.168 + j0.617)
Converting to polar form we have A/B = V'(0.1682 + 0.617:)Z_tan -1
(0.617//0.168) = 0.64/__74.8 ~ Since the real part is negative and the imaginary
part is positive then the phasor falls in the second quadrant so that the angle is
180 - 74.8 = 105.2 ~ from the real positive axis reference.
From Example 4.16 we have that in polar form these two quantities are

A = 7.81/_50.2 ~ and B = 12.21/__-55 ~

Now A//B = Z//B/__(C~z - ~bB) = 7.81//12.21/-[50.2 - (-55)] = 0.64/__105.2 ~

Application to the analysis of series a.c. circuits

The phasor diagram for the series circuit of Fig. 4.14 is given in Fig. 4.15. Note

that the phasor VR lies along the reference axis and that the phasor VL is 90 ~