Alternating Current - Mind Map
Alternating Current - Mind Map
AVERAGE VALUE OF AC FOR ONE TIME PERIOD PHASOR DIAGRAM SERIES AC CIRCUITS
T T
V=√VR2+Vc2 )
I0
I= 0 for 0→T for a sinusoidal ac wave.
VR VL XL l
O
The average value of sin or cos function for I=I0sin(ωt)
l _
tan O=
[
one time period or n time periods (n=1,2...) is zero i R Vc [
v=v0sin(ωt+ )
wt+ l _
tan O=
VR
Keep in mind v=vosinwt V
wt 6. i0= _0
Long period is equivalent to one time period Z 3. i=io sin(ωt+ )
1. V=V0sin ωt VR=i0R, VL=i0X L
l
)O
O R VC V
AVERAGE POWER CONSUMPTION
Mean square current for one time Period vo 2. Voltage phasor diagram 2) R-C CIRCUITS 4. Impedance phasor
T
VL R
∫I2dt
I2 = 0 T =
I02 V=√V +VL2 2
V C R
R
Z = R 2 + X C2
||
[
∫ dt 2 Io VL [
l _
tan O= VR XC
0
Remember VR tan =
The average value of square of sin or cosine VC R
1
2
l
3. i=i0sin (ωt-O) XC Z
function for one time period is 2 1
V= Vosinwt Vo
4. _
Vo = V_ =Z=√R2+XL2 l 5. io =
rms
P= VI
∫T sin2kωt = 1 x T Pavg= Vrms Irms cos io irms )O
0
2 VR Z
for cos =1 or =0o i0
∫nTsin2kωt = n x T
Pavg= Vrms Irms 1. V=V0sin ωt VR=i0 R, VC =i0 X C or irms = Vrms
0
2
Z
3) L-C CIRCUIT RESONANCE IN LCR SERIES CIRCUIT
COMPARISON OF LC OSCILLATION WITH A MASS
QUALITY FACTOR
R1>R2>R3 SPRING SYSTEM
ωr L 1 1 L
i L C In series resonance, impedance of circuit is Q= _ = _ = _ _
R ωr cR R C
minimum & equal to resistance = Z= R, and Mass spring system LC Circuit
[
V = VO sin t
curent is maximum Voltage across C or L [
VL VC or Q= _
~ VL = iO XL , VC = iO XC applied voltage 1.Displacement (x) 1. Charge (q)
Condition for resonance resonance
V=Vosinωt
1 Less sharp the resonance, less is the selectivity 2.Velocity V= _dx 2. Current I= _ dq
dt
Voltage phasor diagram XL = XC L = dt _
C of the circuit.If the Quality factor is large, R dv 3. Rate of change of
V = VL ~ VC [ie, (VL - VC) or (VC - VL)] 1
3. Acceleration a= dt
is low or L is large, the circuit is more selective.
= r = rad / sec
current= (_
dI
4. Mass (m),(inertia)
(
VL XL LC dt
Impedance Phasor Diagram
r resonant frequency (angular) 5. Force constant K 4. Inductance (L), inertia of circuit
Sharpness of Resonance
i
io
1 6. Momentum p = mv 5. Capacitance (C)
io f = fr = ωr 7. Retarding force -m _
dv
Sharpness= Q= _
Hz
Z = XL ~ XC [ (XL - XC) 2 LC ; 2∆ω -bandwidth 6. Magnetic flux O l =LI
2∆ω dt
7. Self induced emf (-L _
VC XC or (XC - XL)] dI (
fr = resonant frequency smaller∆ω, sharper or narrower the 8. Differential equation dt
if XL > Xc , Voltage leads the current by GRAPH
resonance. _
d2x + ω2x=0 8. Differential equation
2
io dt2 d2 q
if XC > XL , current leads the voltage by
2 _ + ω2q=0
if XL = XC , Z = 0, i = POWER IN AC CIRCUIT ω=√ _
k
m
dt2
ω=√ _ 1
w<wr
or, r, XC > XL
i Z w>wr <
ωL = 1 l
Average Power P= Vrms Irms cos O _
1 LC
current leads 9. K.E= 2 mv2 _ LI2
ωC r
Case 1 P= I
2
l
Zcos O 9. Magnetic energy= 1
ω = 1 = ωo Variation of peak current with applied > r, XL > XC rms _
Elastic U =1 kx2 2 2
2 q
_
LC
frequency
current lags l =0
Purely Resistive circuit - O l =1
,COS O Elastic U = 2C
o ωo
In resonance Maximum power dissipation
L-C-R Series Circuit Case 2 TRANSFORMERS
L C R V = VR (applied voltage = voltage across
resistance) Purely inductive or capacitive circuit- “Device which raises or lowers
VL VC VR Z = R (impedance is minimum and equal to l =90
O 0
l
cos O=0 voltage in ac circuits through mutual NP NS
output
Input
resistance) No power is dissipated even though a induction”.Transformer can increase
i V=Vosinωt or decrease voltage or secondary coil
Voltmeter connected across VL & VC will show current is flowing in the circuit current but not both simultaneously.
~ Primary coil
the same reading Case 3
V = VO sin t
Voltmeter connected commonly across inductor LCR Series circuit
VR = iOR, VL = iO XL , VC = iOX C & capacitor shows no reading l non zero in R-L,C-R,or CLR circuit.
O soft iron core
Assuming VL > VC for drawing phasor Here V = 0
P=Vrms Irms Cos O
l EQUATIONS
Voltage phasor diagram V
VL = VC Vs -Voltage in secondary
VL VC VR
V N IP Vp -Voltage in primary
VL
VL - VC VN = VR Case 4 1) _S = _S = _
VP NP IS Ns -No of turns in secondary
VR Power dissipation at resonance Pout V I
io Vnet = VR 2) Efficiency η= _ = _
S S Np -No of turns in primary
VR imax = l
XL-XC=0 or O=0 l
=> cos O=1 ⇒Z=R P in
IVP P Ip -Current in primary
~ Z R
P=I Z = I R
2 2 3) For ideal transformer, η=1 Is -Current in secondary
VC
VR, io Vnet Maximum power is dissipiated in a circuit
at resonance.
V = VR2 + (VL - VC) 2 TRANSFORMER TYPES LOSSES IN TRANSFORMER
Here i = io sin ( t - ) LC OSCILLATIONS step up step down 1) Cu loss (I2R loss)
→ To minimise, windings are made of thick
(since VL is leading ) transformer transformer Cu wires (high resistance)
I
APPLICATION OF RESONANT
Impedance Triangle
CIRCUIT
Capacitor stores 2) Eddy current loss
c → To minimise Cores are laminated
XL - XC XL - XC electrical energy O/P
O/P