Introduction to Electric Circuits

102 Single-phase a.c. circuits

8f~

t!

lOOV

10s

7fl 9fl

f 1

5~

I!

2f~

II

Figure 4.43

Example 4.23

For the series-parallel circuit of Fig. 4.43 determine:

(1) the equivalent impedance of the circuit;

(2) the total power consumed by the circuit;

(3) the reactive power in the capacitive reactance;

(4) the overall power factor of the circuit.

Solution

1 The impedance of the upper inductive branch is

Z 1 --- (8 -+-jl0)1~ = V(8 2 -~- 102)/tan -~ (10/8)1~

SO

Z1 - 12.81/-51.34 ~

The impedance of the lower inductive branch is

Z2- (7 + j9)fl = %/(72 + 92)/_tan -1 (9/7)l-I

SO

Z 2 -- 11.4/_52.13 ~ fl

Now

Z 1 -3 t- Z 2 -- (8 -4- 7) + j(lO + 9)

= (15 + j19) f~ = k/(152 + 192)/_tan -' (19/15) a

so

Z 1 -3 t- Z 2 = 24.2/_51.7 ~ 1)

The equivalent impedance of the parallel inductive branches is