Topic 5 of 7
Describe an alternating supply
An alternating current reverses direction. An alternating voltage reverses polarity relative to named terminals. A negative graph value identifies the opposite direction or polarity to that reference.
A sinusoid is one possible alternating waveform. Its period T is the time for one complete repeat, its frequency f is the number of complete cycles per second, and its peak value x0 is the maximum magnitude of the quantity.
ω = 2πf = 2π/T
T is in seconds, f in hertz (Hz = s-1) and angular frequency ω in rad/s. One cycle is 2π radians. Angular frequency is not numerically the same as frequency in Hz.
For a sine wave whose time origin is an upward zero crossing:
Here x stands for either current or voltage, not a displacement. The argument ωt is in radians. A different starting phase needs a corresponding phase shift in the equation; do not impose an upward zero crossing on a graph that starts elsewhere.
Worked sinusoidal resistor
Keep current and voltage references consistent
A constant 6.00 Ω resistor carries i = 4.00 sin(100πt) A, with t in seconds. Define positive current into the terminal used as positive for the resistor voltage: this is the passive voltage/current reference. Then v = iR.
f = (100π)/(2π) = 50.0 Hz
T = 1/50.0 = 0.0200 s = 20.0 ms
v = 24.0 sin(100πt) V
One period, with separate current and voltage scales
For this constant 6.00 Ω resistor, positive i enters terminal P and v = VP - VQ. Thus v = Ri and the quantities have the same phase. Time zero is their upward zero crossing; the horizontal scale is identical in both graphs.
Current reverses direction
Voltage reverses with the current
The peaks are 4.00 A and 24.0 V; peak-to-peak values are twice these. The common period gives f = 50.0 Hz. A negative current reading means current flows from Q to P.
The current peaks are +4.00 and -4.00 A, giving peak magnitude 4.00 A and peak-to-peak range 8.00 A. Voltage peak magnitude is 24.0 V and its peak-to-peak range is 48.0 V.
At 5.00 ms the current reaches +4.00 A; at 10.0 ms it crosses zero downward; at 15.0 ms it reaches -4.00 A; at 20.0 ms it returns to the initial upward crossing. A positive maximum and the next negative minimum are only half a cycle apart.
For an intermediate value, convert 2.50 ms to 0.00250 s before substitution:
= 4.00 sin(π/4) = 2.83 A
v = iR = 12√2 V ≈ 17.0 V
The graphs are smooth sinusoids, not straight lines between a few labelled times. The instantaneous values depend on phase. They are not fixed effective values for the whole cycle.
Read a measured or supplied trace
Identify the zero reference, voltage or current scale and time-base scale. Measure the time between equivalent points moving in the same sense, such as two upward zero crossings. Measuring several complete cycles can reduce the fractional error in timing, provided the frequency remains stable. Measure peak magnitude from zero, not from the negative trough.
A suitable low-voltage source and waveform instrument, or a supplied trace, can show the pattern. Preserve the instrument settings and stated calibration with measured readings. A meter's a.c. display must be interpreted using its specified waveform and rms response; not every a.c. meter gives a true rms reading for an arbitrary waveform.
The rms value describes equivalent heating over a cycle. It is different from a peak, a signed mean or a chosen instantaneous reading.