Lecture 20

# Lecture 20 - AC voltage sources As we have shown the use of...

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AC voltage sources As we have shown, the use of rotating elements in generators and the ability to use transformers led to the id l d ti f AC lt widescale adoption of AC voltage sources. For these, the voltage varies with time as V(t)=V 0 cos(2 π ft)= V 0 cos( ω t). V(t) V ft) V V 0 is the amplitude. In common US household applications V 0 170 V. Huh? What about 120 V? In E th l V i 311 V Europe, the analogous V 0 is 311 V. f is the frequency. In the USA, f=60 Hz. Europe uses 50 Hz. This was an unlucky choice as voltages at f 60 Hz Hz. This was an unlucky choice as voltages at f 60 Hz present very high relative shock hazards. It’s too late to change.

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A resistor connected to AC coltage Notice we complete the circuit with ground connections. We have V 0 cos(2 π ft)-i(t)R=0, so i(t)=(V 0 /R)cos(2 π ft). The current is in phase with the voltage. That is, if we expressed i(t) as i(t)=i 0 cos(2 π ft- φ ), with φ the phase constant we have φ =0 for a resistor , we have =0 for a resistor. The power delivered is P(t)=V(t)i(t)=V 0 2 cos 2 (2 π ft)/R. The power is not constant, but the fast oscillation is often not important important. Of more interest is the average power.
Average power and RMS voltage The average power is <P>=<V 0 cos(2 π ft)i 0 cos(2 π ft- φ )>, where <x> means the “time average of x”, and to be general we let the current has a phase constant. This can be written <P>=V 0 i 0 <cos 2 (2 π ft)cos φ +cos(2 π ft)sin(2 π ft)sin φ > <P> V <cos (2 > =(V 0 i 0 /2)cos φ , So <P>=V 0 2 /2R for a resistive load. It is conventional to write this as <P>=V 2 /R where V [<V 2 >] 1/2 =V / 2 rms /R, where V rms =V 0 / 2. It is V rms that equals 120V in USA household electricity applications, and 220V in European household l i i electricity. One can similarly define i rms =[<i 2 >] 1/2 , so in general <P> = V rms i rms cos φ . rms rms The number cos φ is sometimes called the power factor . It is 1 for a resistor.

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An inductor connected to AC coltage N V (2 ft) Ldi/dt 0 i(t) (V /2 fL) i (2 ft) F Now V 0 cos(2 π ft)-Ldi/dt=0, so i(t)=(V 0 /2 π fL)sin(2 π ft). For simplicity, we have taken φ =0.
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