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Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly change. Mains frequency is 50 Hz

Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly

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Page 1: Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly

Alternating Current

voltage

time0

maximum positive voltage (the peak voltage)

maximum negative voltage (peak voltage)

Direction of current and size constantly change. Mains frequency is 50 Hz

Page 2: Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly

Alternating Current 2

Voltage(V)

Time (s)

0

peak voltage

effective voltage

The quoted value ( rms ) or Effective voltage of an a.c. supply is the same as the steady d.c. voltage which would produce the same heating effect in a resistance wire.

The quoted value of the mains voltage in the UK is 230 V. The peak voltage is 325 V.

Effective voltage is measured with a voltmeter. Peak voltage is measured with an oscilloscope.

Page 3: Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly

Alternating Current 3The quoted value of an a.c. supply is also known as the r.m.s. value (root mean square). The peak value is √2 larger than this r.m.s. value. This applies to both voltage and current.

peaksmrpeaksmr IIandVV2

1

2

1......

Voltage

Time

peak value

r.m.s value

Page 4: Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly

Alternating Current 4

2peak

rms

II

AI rms 83.2

Peak and r.m.s. values: Example The period of a sinusoidally varying current is 0.005 s. The peak value of the current is 4·0 A.

Calculate a) the frequencyb) the root mean square value of the current.

Frequency = 1 / Period

Frequency = 1 / 0.005

Frequency = 200 Hz 2

4rmsI

Page 5: Alternating Current voltage time 0 maximum positive voltage (the peak voltage) maximum negative voltage (peak voltage) Direction of current and size constantly

Calculation of frequency

timebase

(s cm –1 )

y input0.02

Find the time for one wave.

This is the period.

frequency = 1/period

ExampleTime base = 0.02 s cm-1

Period = time base setting x no of boxes for 1 cycle

Period = 0.002 x 4 = 0.08 s

Frequency = 1 / period

Frequency = 1 / 0.08 = 12.5 Hz.