Introduction to Electric Circuits

4.3 Series a.c. circuits 77

Solution

The capacitive reactance Xc = VII = 100/5 = 20 f~. From Xc = 1/2zrfC we see

that the frequency f = 1/2zrCXc = 1/(27r 120 x 10 -6 x 20), so./= 66.3 Hz.

Summary

For the single-element a.c. circuits, which can also be considered as building

blocks for the multiple-element circuits which follow, we have seen that"

For the pure resistor, the current is in phase with the source voltage and

V = IR (4.12)

For the pure inductor, the current lags the source voltage by 90 ~ (zr/2 radians)

and V = IXL (4.13)

For the pure capacitor, the current leads the source voltage by 90 ~ (z r/2

radians) and V = IXc (4.14)

The inductive reactance XL = 2zrfL (4.15)

The capacitive reactance Xc = 1/2zrfC (4.16)

4.3 SERIES A.C. CIRCUITS

Series RL circuits

In practice resistive circuits will have some inductance however small because
the circuit must contain at least one loop of connecting wire. Also inductive
circuits must have some resistance due to the resistance of the wire making up
the coil. It is usual to show the resistance of a coil as a separate pure lumped
resistor in series with a pure inductor as shown in Fig. 4.14. We can then make
use of our building blocks assembled above.

Figure 4.14

I R L

I I

~VR VL

tO

l

Kirchhoff's laws can be applied to a.c. circuits in the same way as for d.c.
circuits provided we use phasor sums rather than algebraic sums. Applying