
R–L–C Circuits and Resonance 149
Example 4.1 A resistance of 10 Ω, a capacitor of 470 pF and inductor of 0.5 mH are connected
in series across a variable frequency voltage source. Calculate the value of frequency at which
the circuit will attain the resonance condition.
Solution:
f
LC
r
2
==
××××
2314 05 10 470 10
2
p
..
4.2.1 Effect of Variation of Frequency on Current
and Voltage Drops
In a series R–L–C circuit, at a frequency f
r
, X
L
= X
C
. Below this resonant frequency, X
L
is less
than X
C
as frequency is decreased from f
r
, X
L
decreases but X
C
increases (X
L
is directly propor-
tional to frequency, while X
C
is inversely proportional to frequency). ...