Introduction to Electric Circuits

Example 4.12

4.4 Complex notation 87

Determine the sum of the two voltages V 1 = (10 + j50) g and
V2 = (15 -j25) V.

Solution

Let the resultant voltage be V. Then

V = (10 + 15) + j(50 - 25) = (25 + j25) V

The magnitude of V is given by ~V/(252 + 252) -- 35.35 V. The phase of V with
respect to the reference is given by 4,- tan -~ (25/25)- 45 ~ The complete
phasor diagram is shown in Fig. 4.26.

Figure 4.26

Example 4.13

Two currents entering a node in a circuit are given by 11 = 20/_30~ A and

12 = 30/45 ~ A. Calculate the magnitude and phase of a third current 13 leaving

the node.

Solution

Converting to rectangular coordinates we have

I1 - 20 cos 30 ~ + j20 sin 30 ~ = (17.32 + jl0) A

I: = 30 cos 45 ~ + j30 sin 45 ~ - (21.2 + j21.2) A

By Kirchhoff's current law,

I3 = I1 + I2 - (17.32 + 21.2) + j(10 + 21.2) = (38.52 + j31.2) A

I3 = S/(38.522 + 31.22)/-tan -1 (31.2/38.52)= 49.57/_39 ~ A