Hello friends in this video we are going
to see a problem based on series
resonance the statement of problem is
like this a series RLC circuit has the
following parameter values R equal to
10 ohm L equal to 0.01 for Henry and C
equal to 100 micro farad compute the
following first resonant frequency in
Radian per second second quality factor
of the circuit third bandwidth fourth lower
and upper frequency points of the
bandwidth and finally maximum values of
the voltage appearing across the
capacitor if the voltage V of T equal to
1 sin 1000 T is applied to the RLC
circuit so lets solve clearly it is a
problem of series resonance where value
of R is 10 ohm L 0.014 Henry and C
100 micro farad so lets calculate
resonant frequency they ask in Radian
per second so Omega R for series
resonance will be 1 upon root LC so this
is equal to 1 upon root 0.014 multiplied
by 100 multiplied by 10 raised to minus
6 because capacitor they have given 100
micro farad so if we solve will get
Omega R equal to 845.1543
 Radian per second
the first answer we got simply by using
the formula second quality factor of the
circuit so quality factor of the circuit
is given as Q equal to 1 by R root of L
by c so if i substitute
it is 1 by 10 root of 0.014 divided by
100 multiplied by 10 raised to minus 6
if you solve you will get quality factor
as 1.1832 its a unit-less
quantity because it is just a ratio
third third part we have to get a
bandwidth so bandwidth is given as R by
L so it is 10 divided by 0.014 so
answer we will get equal to 
714.2857 Radian
per second and in fourth part we are
supposed to get lower and upper cutoff
frequency points so for that I will draw
simple diagram like this
so at resonance I am getting current
maximum and these are the points where I
am getting current 0.707 times Im
current bonding frequencies are nothing
but FL and FU or you can say FL is
nothing but lower cutoff frequency and
FH is higher cutoff frequency and this
distance is nothing but bandwidth so I
can say bandwidth is nothing but FH
minus FL but if I am talking in terms of
Hertz as a unit of frequency then this
formulas are applicable otherwise it
will be like this bandwidth I can say it
is Omega U minus Omega L and how to get
Omega U and Omega L so Omega L I can say
from this it is resonating frequency
minus half of the bandwidth Omega L
equal to Omega R minus bandwidth divided
by 2 so if I substitute what I will get
omega L as 845.1543
 minus 714.2857
 divided by
2 so after solving this I will get
lower cutoff frequency equal to 488.0114
 Radian per
second
similarly upper cutoff frequency will be
what resonating frequency plus half of
the bandwidth so upper cutoff frequency
equal to Omega R plus bandwidth divided
by 2 so this is equal to 845.1543
plus
714.2857
 divided by 2 so if we solve you
will get upper cutoff frequency equal to
1202.2972
Radian per second lets go to the last
part in last part what they have said
what is the maximum value of voltage
appearing across the capacitor if the
voltage V of T equal to 1 sin 1000
T is applied to the RLC circuit so what
they are saying in the last part I have
a voltage applied to the circuit equal
to 1 sin 1000 T what will remain
same quality factor so Quality factor we
obtain as 1.1832
1.1832 so what
they are saying is quality factor is
given as voltage across capacitor or
voltage across inductor at resonance
divided by
supplied voltage VS we are talking in
terms of all RMS values over here so I
can say maximum value of voltage across
capacitor simply equal to its quality
factor multiplied by VS so VC max will
be equal to 1.1832
multiplied by 1 because here if you
compare with the standard equation VS
maximum you will get as 1 volt so if I
solve maximum voltage across capacitor
will be equal to 1.1832
volt which is the required
answer thank you
