alternating current

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