Introduction to Electric Circuits

Example 4.18

4.5 Parallel a.c. circuits 91

A coil having a resistance of 2.5 1) and the inductance of 60 mH is connected in

series with a capacitor having a capacitance of 6.8 txF to a 230 V, 50 Hz supply.

Determine the current drawn from the supply

Solution

I 2.5fl 60mH 6.81~F

230v

Figure 4.28

The circuit is shown in Fig. 4.28. Take the voltage as the reference so that
V = 230/_0 ~ V. The impedance of the circuit is given by Z = R + j(XL- Xc).
Now

X L -- 2-rrfL = 2rr50 x 60 • 10 -3= 18.85

and

Xc = 1/27rfC = 1/(27r50 • 6.8 • 10 -6) = 468 l)

Thus XL -- Xc = -449.15 12 and Z = (2.5 - j 449.15) ft. In polar form

Z = ~/(2.52 + 449.152)/_tan -1 (449.15/2.5)= 449.2/_-89.42 ~ l)

The current is
I = V/Z = 230/00/449.2/-89.42 ~ 230/449.2/[0 ~ (-89.42~
so I = 0.512/__89.42 ~ A. The current therefore leads the voltage as expected in a
predominantly capacitive circuit.

4.5 PARALLELA.C. CIRCUITS

Parallel RL circuits

The circuit on the following page of Fig. 4.29(a) shows a resistor R in parallel
with an inductor L. The corresponding phasor diagram is given in Fig. 4.29(b).
It is convenient to take the voltage V as the reference phasor since it is common
to both elements.
The current I R is in phase with the voltage and the current Ic is 90 ~ behind
(lagging) the voltage. Kirchhoff's current law tells us that the total current I is
the phasor sum of I R and I L. The magnitude of I is V(IR 2 + IL 2) and the phase
angle ~ is given by tan -~ (IL/IR). Since I = V/Z, IR = V/R and IL = V/XL

V/Z = V'[(V/R) 2 + (V/XL) 2] = VX/[(1/R) 2 + (1/XL) 2]