Analog Electronics

Passive circuits

37

you felt

so

inclined. Then becomes

α

(say)

and

7:

= 1/a.

Resonant circuits

Figure

2.7

shows

an

important example

of a

two-pole (second-order) circuit.

At

some fre-

quency

the

circuit will

be

resonant

i.e. |]ω7| =

1ΐ/]ωΓ|. At

this frequency,

the PD

across

the

inductor will

be

equal

in

magnitude

and

opposite

in sign

to the PD

across

the

capacitor,

so

that

the

net

PD

across

L and C

together

(but not

across

each separately) will

be

zero.

The

current

ζ

will

then exhibit

a

maximum

of / = vJR. At

resonance,

then,

ΐ[]ωΕ

+

(1/]ωΟ]

=

O,

so ω-LC = 1 and

ω

= 1/V(LC).

Let's give this value

of ω the

label

ω„.

At ω,, the

output voltage