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lf1206 lf1206
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Posts: 16
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12 years ago
resistor voltage(peak to peak)=8.4V
phase angle?=42.4degrees (i think it means phase angle what else could phase mean).

its in a series circuit with capacitor=0.18uF, 1Kohm resistor,
function generator(12V(p-p)/1Khz/0degree phase)
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wrote...
12 years ago
divide the voltage of the resistor by 1K to find the current in the circuit:

i=4.2<41.4 A (leading)

p.f=cos (-41.4)
wrote...
12 years ago
Using given resistor Voltage and 42.4 degree phase angle
Circuit current = 8.4Vpp/1,000 Ohm = .0084 Amps (peak to peak)= 8.4mA (peak to peak)
Power factor = cos 42.4 degrees = .74

But some of the given values must be incorrect because:
Xc at 1kHz = 1/2pi fc = 1/[(6.28) x (1,000 Ohm) x (.00000018 F) = 884.64 Ohm
The real phase angle at 1kHz = arc tan Xc/R = arc tan (884.64 Ohm) / (1,000 Ohm) = 41.5 degrees
Power factor = cos of phase angle = cos 41.5 degrees = .75
Circuit Z = 1,000 Ohm / cos 41.5 degree = 1335.19 ohm
Circuit current = 12v p-p / 1335.19 Ohm = .00899 Amps = 8.99mA
Resistor Voltage (peak to peak) = 8.99mA x 1,000 Ohm = 8.99 Volts
Julie Appleton
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