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NEOCLASSICAL
Alternating Current
NEOCLASSICAL
Alternating Current Waveforms
NEOCLASSICAL
RMS Value
RMS Root Mean square
First square
Then Take Mean
Then Take Square root
Mean of sin =0
Mean of sin2 = ½ ( because sin2 = ½-cos 2x/2 , and average of cos2x= 0 , only ½
Mean of Constant , K = K
NEOCLASSICAL
For Sinusoidal waves only› Vrms =
› Vaverage = 0
› Average of sinusoid is Zero
NEOCLASSICAL
› Root Mean Square of Sinusoid
› V= Vpeak cos ( w t + Ø)
› First step : Square V 2peak cos 2( w t + Ø)
› Second step : Mean : V 2peak / 2
› Third step : Root =
NEOCLASSICAL
V = 3 + 4 cos (wt + 4)
Vaverage = Average (3) + Average (4 cos (wt + 4))
= 3 + 0 = 3
V rms
1 . Square = 9 + 16 cos 2(wt + 4 ) + 24 cos (wt + 4)
2 . Mean = Mean(9) + Mean(16 cos 2(wt + 4 ) ) +Mean(24 cos (wt + 4)
= 9 + (16/2) + 0 = 17
3 Root :
NEOCLASSICAL
Resistor V= Vpeak cos (ωt +φ)
I = cos (ωt +φ)
V I
Current and Voltage are in phase for resistor
NEOCLASSICAL
Inductor
› V= Vpeak cos (ωt +φ)
Reactance of Inductor = XL=ωL
I = cos (ωt +φ- )
Current lags voltage by π/2
V
I
NEOCLASSICAL
Voltage across inductor of 1 Henry = 200 cos (100t +φ), Find current
w = 200 L = 1 Reactance = 200
Current peak = Vpeak / Reactance = 200/200 =1
Now current will be π/2 behind voltage
So , I =1 cos (100t +φ – π/2),
If current is given
Vpeak = Ipeak X Reactance = 1 X 200 =200
Current is π/2 behind voltage voltage is π/2 ahead
V = 200 cos (100t +φ – π/2+ π/2)= 200 cos (100t +φ )
NEOCLASSICAL
Capacitor
› V= Vpeak cos (ωt +φ)
Reactance of Capacitor= Xc=
I = (ωC)Vpeak cos (ωt +φ+ )
Current leads voltage by π/2 V
I
NEOCLASSICAL
GENERAL RLC CIRCUIT
› Input : V= Vpeak cos (ωt +φ)
Current will have same form
Except amplitude and phase
I = Ipeak cos (ωt +φ + φextra)
This current will pass through all three as they are in series
NEOCLASSICAL
› Magnitude
› Phase, φextra
› Φextra will be negative if XL > X c
NEOCLASSICAL
I peak = φextra =
I = Ipeak cos (ωt +φ + φextra)
If Current is given
Vpeak= Ipeak X Z
Find φextra and subtract
NEOCLASSICAL
Power in RLC
› Power = cos(φextra )
= V rms Irms cos(φextra )
Power is lost only through Resistor
NEOCLASSICAL
Resonance› At resonance Amplitude is Max Z is minimum
Frequency = ω / 2 π
NEOCLASSICAL
Quality Factor
Greater the Q , sharper is the Resonance
(i) is sharper