ELECTRICAL POWER resistAnCe, induCtAnCe And CApACitAnCe on AC
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Electrical Power Summary
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RESISTANCE, INDUCTANCE AND
ELECTRICAL POWER resistAnCe, induCtAnCe And CApACitAnCe on AC
Electrical Power Summary
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The current lags the voltage by 1/4 cycle
Figure 61;Current in an inductor
AC Current through a Capacitance (or Condenser) Let the capacitance be C farads
Thus the charge q=C v where dq/dt=i amps q = C Vp Sin 2πf t Differentiating wrtt
i = C Vp 2πf Cos 2πf t Hence Ip = 2πf C Vp
Irms=Ip/√2 and Vrms=Vp/√2 Hence Irms = 2πf CVrms
But Vrms= IrmsXC
where XC is the reactance of the capacitor in ohms (ie volts/amps) Thus XC = 1/(2πf C)
WithCin F then XC = 106/(2πf C) At 50 Hz XC = 3183/C
whereXC is the reactance of the capacitor in ohms (ie volts/amps) and Cis in F
Figure 62;Current leads the voltage
Figure 63;Current in a Capacitance
The current is established over several cycles, but it then lags by 3/4 cycle.
Thus, in effect, Capacitive Current leads the voltage by 1/4 cycle
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ELECTRICAL POWER resistAnCe, induCtAnCe And CApACitAnCe on AC
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Eddy Currents in a conductor
AC Resistance is higher than DC Resistance due to eddy current loss.
A conductor, radius a metres carries a currentISin(pt) amps
Figure 64;Current in a conductor
Houtside the conductor is (2I/r) Sin (pt) 10–7
whereris the distance in metres from the centre of the conductor At the surface of the conductor
H= (2I/a) Sin (pt) 10–7
At the centre of the conductor,H= 0 Thus inside the conductor
H= (2I r/a2) Sin (pt) 10–7
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ELECTRICAL POWERElectrical Power resistAnCe, induCtAnCe And CApACitAnCe on ACSummary
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Letσbe the eddy current in amps/m2at radiusrmetres
andσ0be the current in amps/m2at the centre of the conductor (σ+σ0)ρ=─ ∂Φ/ ∂t
=─ ∫ [ ∂H / ∂t] dr fromr= 0 tor=r
=─ ∫ [ (2 p I / a2) Cos (pt) ] 10–7rdr fromr= 0 tor=r
= (p I r2/a2) Cos (pt) 10–7
therefore σ= {p I r2/ (ρa2)} Cos (pt) 10–7─σ0
But 2π ∫ σrdr= 0 fromr= 0 tor=a
2π ∫ [ {p I r2/ (ρa2)} Cos (pt) 10–7─σ0]rdr= 0 fromr= 0 tor=a 2π [ {p I r4/ (4ρa2)} Cos (pt) 10–7─σ0r2/2] = 0 fromr= 0 tor=a p I a2/ (4ρ) Cos (pt) 10–7 ─σ0a2/2] = 0
σ0=p I/ (2ρ) Cos (pt) 10–7
therefore σ= (p I/ρ) Cos (pt) [r2/a2 ─½]10–7 andσ= 0 whenr=a/√2 Total current density i=σ+ [ISin (pt) ] / (π a2)
Put [ISin (pt) ] / (π a2) =C
i2=σ2 + 2σC+C2
Energy loss = 2π ∫ i2 ρrdr fromr= 0 tor= a
= 2π ∫ [σ2 + 2σC+C2]ρrdr fromr= 0 tor=a But as before, 2π ∫ σrdr= 0 fromr= 0 tor=a Therefore 2π ∫ 2 σCρrdr= 0 fromr= 0 tor=a Energy loss = 2π ∫ [σ2+C2]ρrdr fromr= 0 tor=a Additional loss due to eddy currents = integral from 0 toa
2π ∫ σ2ρrdr = 2π (p2I2/ρ) Cos2(pt) 10–14 ∫ [r5/a4 ─r3/a2 +r/4] dr
= 2π (p2I2/ρ) Cos2(pt) [a2/6 ─a2/4 +a2/8] 10–14
= π p2I2a2/(12ρ) Cos2(pt) 10–14 Mean value ofImax2Cos2(pt) =Irms2
Mean value of energy loss = π p2Irms2a2/(12ρ) 10–14 Total loss =Irms2[ρ/ (π a2 )+ π p2a2/(12ρ) 10–14]
=Irms2R0 [ 1 + π2p2a4/(12ρ2) 10–14] whereais in metres andρis in ohms per metre cube hence Rf /R0 = 1 + π2p2a4/(12ρ2) 10–14
butp = 2πf
Rf /R0 = 1 + 100π4f2a4/(3ρ2)
wherefis in Hz,ais in metres andρis inμΩper cm cube Example
Conductor 1.29 cm radius, 50 Hz,ρ= 1.65μΩper cm cube Rf / R0= 1 +1003.144502(1.29/100)4 /(31.652)
= 1.08 Iron cored inductor
The iron core of an inductor can saturate as the sine wave approaches peak value. If an AC voltage is applied, the current will increase as the peak value is approached. The current wave has a peaky form factor.
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Figure 65;Current in an inductor
There is another effect. If the rms current through an iron cored reactor is increased, the reactance falls sharply as the iron saturates. If the current is increased in stages from zero and the voltage drop across the reactor is measured, the graph will have a kink as shown. This means that applying a fixed voltage Vc to the reactor, the current can take any one of three values. One of these values, I2 in the diagram, is unstable but the other two are stable. So while applying a voltage within this range, any transient can make the current flip to the other value.
Figure 66;Current in an inductance
Examples Find the current taken from the 240 volt 50 Hz mains by;
(i) 0.12 henry inductance (ii)40μF capacitance.
(i)X= 3140.12 = 37.7 ohms I=V/X= 6.37 amps (ii)X= 3183/40 = 79.6 ohms I=V/X= 3.02 amps
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ELECTRICAL POWERElectrical Power AC CirCuitsSummary
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