Introduction to Electric Circuits

6. 1 Series resonance 127

Zmi n - _ I
I I
0 fo
Figure 6.7

I

|max

:~ 0 fo

Figure 6.8

care must therefore be taken when dealing with series circuits containing

inductance and capacitance.

Example 6.4

A coil having a resistance of 5 f~ and inductance of 10 mH is connected in series

with a capacitor of 250 nF to a variable frequency, 100 V supply. Determine the

potential difference across the capacitor at resonance.

Solution

Figure 6.9

oovl@

5s lOmH 250nF

,f t tl

The circuit is shown in Fig. 6.9. Using Equation (6.2)

fo = 1/2~'V'[10 • 10-3x 250 • 10-91 - 3183 Hz

Xc = 1/2"rrf0C = 1/2vr • 31.83 • 250 • 10 -9- 200 a

The current at resonance is limited only by the resistance and

I = V/R = 100/5 - 20 A. The potential difference across the capacitor is given

by Vc = IXc = 20 • 200 = 4000 V. Note that this is 40 times greater than the

supply voltage V.

Example 6.5

A 20 mH coil has a resistance of 50 1) and is connected in series with a capacitor

to a 250 mV supply. If the circuit is to resonate at 100 kHz calculate the

capacitance of the capacitor and its working voltage.