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Electrical Circuit II
EEE-121
Md. Mokarram Hossain Chowdhury
Lecturer
Department of Electrical & Electronic Engineering
Stamford University Bangladesh
Resonance In Electric Circuits
 Any passive electric circuit will resonate if it
has an inductor and capacitor.
 Resonance is characterized by the input
voltage and currentbeing in phase. The driving
point impedance (or admittance) is completely
real when this condition exists.
 In this presentation we will consider (a) series
resonance, and (b) parallel resonance.
Series Resonance
Consider the series RLC circuit shown below
V = VM 0
V
The input impedance is given by:
The magnitude of the circuit current is;
1
( )Z R j wL
wC
  
2 2
| |
1
( )
mV
I I
R wL
wC
 
 
Series Resonance
Resonance occurs when
2πfrL =
1
2π𝑓𝑟𝐶
At resonance we designate w as wo and write
This is an important equation to remember. It
applies to both series And parallel resonant
circuits.
1
wL
wC

1
ow
LC

Series Resonance
Series Resonance
Series Resonance
Series Resonance
 The magnitude of the current response for the
series resonance circuit is as shown below.
Half power point
Bandwidth:
BW = wBW = w2 – w1
mV
R
2
mV
R
Series Resonance
The peak power delivered to the circuit is;
The so-called half-power is given when
We find the frequencies, w1 and w2, at which this
half-power occurs by using;
2
mV
P
R

2
mV
I
R

2 21
2 ( )R R wL
wC
  
Series Resonance
After some insightful algebra one will find two
frequencies at which the previous equation is
satisfied, they are:
And
The two half-power frequencies are related to the
resonant frequency by
2
1
1
2 2
R R
w
L L LC
 
    
 
2
2
1
2 2
R R
w
L L LC
 
   
 
1 2ow w w
Series Resonance
The bandwidth of the series resonant circuit is given
by;
We define the Q (quality factor) of the circuit as;
Using Q, we can write the bandwidth as;
These are all important relationships
2 1b
R
BW w w w
L
   
1 1o
o
w L L
Q
R w RC R C
 
    
 
ow
BW
Q

Series Resonance
Series Resonance
Q-factor: The quality factor of RLC circuit is the
voltage magnification in the circuit at resonance.
Voltage Magnification =
𝑉𝐿
𝑉
or
𝑉𝑐
𝑉
Parallel Resonance
 Circuit Current is in phase with the voltage applied
 Phase angle Zero
 Power factor cosØ = 1







jwL
jwC
R
VI
11
Parallel Resonance
 Zrl=R1+jXL
 Y= 1/Zrl
 Y= 1/R1+jXL
 Y= 1/R1+jXL * R1-jXL/R1-jXL
 Y=R1-jXL/R12+XL2
 *****
Parallel Resonance
Determine the resonant frequency for the circuit
below
jwRCLCw
jwLLRCw
jwC
jwLR
jwC
RjwL
Z NI






)1(
)(
1
)
1
(
2
2
At resonance, the phase angle of Z
must be equal to zero.
Resonance
jwRCLCw
jwLLRCw


)1(
)(
2
2
LCw
wRC
LCRw
wL
22
1()( 


12222
 CRwLCw
)(
1
22
CRLC
wo


Analysis
For zero phase;
This gives;
or
Parallel Resonance
A series RLC resonant circuit has a resonant
frequency admittance of 2x10-2 S(mohs). The Q
of the circuit is 50, and the resonant frequency
is10,000 rad/sec. Calculate the values of R, L,
and C. Find the half-power frequencies and the
bandwidth.
Thank you
for your kind attention
Question ?

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Electrical circuit ii (eee 121)

  • 1. Electrical Circuit II EEE-121 Md. Mokarram Hossain Chowdhury Lecturer Department of Electrical & Electronic Engineering Stamford University Bangladesh
  • 2. Resonance In Electric Circuits  Any passive electric circuit will resonate if it has an inductor and capacitor.  Resonance is characterized by the input voltage and currentbeing in phase. The driving point impedance (or admittance) is completely real when this condition exists.  In this presentation we will consider (a) series resonance, and (b) parallel resonance.
  • 3. Series Resonance Consider the series RLC circuit shown below V = VM 0 V The input impedance is given by: The magnitude of the circuit current is; 1 ( )Z R j wL wC    2 2 | | 1 ( ) mV I I R wL wC    
  • 4. Series Resonance Resonance occurs when 2πfrL = 1 2π𝑓𝑟𝐶 At resonance we designate w as wo and write This is an important equation to remember. It applies to both series And parallel resonant circuits. 1 wL wC  1 ow LC 
  • 8. Series Resonance  The magnitude of the current response for the series resonance circuit is as shown below. Half power point Bandwidth: BW = wBW = w2 – w1 mV R 2 mV R
  • 9. Series Resonance The peak power delivered to the circuit is; The so-called half-power is given when We find the frequencies, w1 and w2, at which this half-power occurs by using; 2 mV P R  2 mV I R  2 21 2 ( )R R wL wC   
  • 10. Series Resonance After some insightful algebra one will find two frequencies at which the previous equation is satisfied, they are: And The two half-power frequencies are related to the resonant frequency by 2 1 1 2 2 R R w L L LC          2 2 1 2 2 R R w L L LC         1 2ow w w
  • 11. Series Resonance The bandwidth of the series resonant circuit is given by; We define the Q (quality factor) of the circuit as; Using Q, we can write the bandwidth as; These are all important relationships 2 1b R BW w w w L     1 1o o w L L Q R w RC R C          ow BW Q 
  • 13. Series Resonance Q-factor: The quality factor of RLC circuit is the voltage magnification in the circuit at resonance. Voltage Magnification = 𝑉𝐿 𝑉 or 𝑉𝑐 𝑉
  • 14. Parallel Resonance  Circuit Current is in phase with the voltage applied  Phase angle Zero  Power factor cosØ = 1        jwL jwC R VI 11
  • 15. Parallel Resonance  Zrl=R1+jXL  Y= 1/Zrl  Y= 1/R1+jXL  Y= 1/R1+jXL * R1-jXL/R1-jXL  Y=R1-jXL/R12+XL2  *****
  • 16. Parallel Resonance Determine the resonant frequency for the circuit below jwRCLCw jwLLRCw jwC jwLR jwC RjwL Z NI       )1( )( 1 ) 1 ( 2 2 At resonance, the phase angle of Z must be equal to zero.
  • 18. Parallel Resonance A series RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth.
  • 19. Thank you for your kind attention Question ?