From your description, this sounds like it is a sine wave offset to 10A, so the peak is at 20A, and the min is at 0?
For this case, you have 10A DC (RMS) wave and a 10A Peak - neutral AC wave; The RMS value of the AC wave is: 10/2*sqrt(2) = 3.54A.
So the RMS amplitude of this wave is 13.54A.
The root-mean-square value is 0.707 times the peak value, for a sinusoidal voltage or current. Angle doesn't come into it.
From your description, this sounds like it is a sine wave offset to 10A, so the peak is at 20A, and the min is at 0? For this case, you have 10A DC (RMS) wave and a 10A Peak - neutral AC wave; The RMS value of the AC wave is: 10/2*sqrt(2) = 3.54A. So the RMS amplitude of this wave is 13.54A.
One is sinusoidal through the zero point i.e. alternating current, the other would be a flat line at whatever the current value was i.e. direct current
AC voltage, like the voltage in your house, is typically referred to as 120vAC. This means the voltage swings 120V positive and 120V negative 60 times per second (60 Hz) 360 degrees total (sine wave). Current and voltage go hand-n-hand so the current alternates with the voltage. the RMS value is what we experience at the output (160vAC is actually sent to the circuits)
Unless otherwise stated, the value of an a.c. current or voltage is expressed in r.m.s. (root mean square) values which, for a sinusoidal waveform, is 0.707 times their peak value. The output of a voltage (or potential) transformer is no different, its measured voltage will be its r.m.s value which is lower than its peak value.
The root-mean-square value is 0.707 times the peak value, for a sinusoidal voltage or current. Angle doesn't come into it.
From your description, this sounds like it is a sine wave offset to 10A, so the peak is at 20A, and the min is at 0? For this case, you have 10A DC (RMS) wave and a 10A Peak - neutral AC wave; The RMS value of the AC wave is: 10/2*sqrt(2) = 3.54A. So the RMS amplitude of this wave is 13.54A.
If you are referring to an a.c. current, then the maximum current is the amplitude of its waveform. For a sinusoidal waveform, the amplitude of an a.c. current is its root-mean-square value, divided by 0.707. For example, an a.c current with an rms value of, say, 10 A will have an amplitude of 14.14 A,
Presumably you are referring to an a.c. current?If so, then the average value of an a.c. current is zero so, clearly, you cannot determine its maximum value.However, average current is more-usually applied over half a cycle, in which case, for a sinusoidal current, this value is 0.637 Imax. So the maximum current will be the average value, divided by 0.637.
One is sinusoidal through the zero point i.e. alternating current, the other would be a flat line at whatever the current value was i.e. direct current
Both of them have the same value, ie, Im/2. Where Im is the max current through the load. NOTE: This data is for the sinusoidal input.
The root-mean-square (rms) value of a sinusoidal voltage or current is given by: Vrms = 0.707 Vmax and Irms = 0.707 ImaxSo, if the current has a peak-to-peak value of 10 A, then Imax will be half that value (5 A) , so the corresponding rms value is:Irms = 0.707 Imax = 0.707 x 5 = 3.54 A(Answer)
If an AC current is sinusoidal, then the peak value of a current with an RMS value of 2A is 2.8A, or 2A times the square root of 2.However, very few current waveforms are sinusoidal. In fact, they are often far from sinusoidal - they are more often pulsating - because most power supplies, even switching power supplies, only draw current when the rectifier diodes conduct to recharge the primary filter capacitor. That only occurs once (half wave rectifier) or twice (full wave rectifier) each line cycle.The square root of 2 rule does apply if the load is purely resistive and the voltage is also sinusoidal. Most voltages are also not sinudoidal, because of the pulsating current issue, and because conductors are not perfect zero impedance conductors. Voltage waveforms, however, are more closely sinudoidal than current waveforms.The most accurate way to measure the peak current value is to observe it with an oscilloscope. If you want the analytic approach, you need to back calculate from peak value to RMS value, knowing that RMS value is the square root of the sum of the squares of the observations divided by N, as delta T approaches zero, and N approaches infinity. (Best to just use the oscilloscope.)
Hi, RMS is voltage X .707 and the power is voltage X current. Hope that helps, Cubby
You are, presumably, referring to alternating current, in which case the 'maximum' current is the peak or amplitude of the waveform. The 'average' value of current is zero, because the average value of the first half of each cycle is negated by the average value over the second half of each cycle. This is why a.c. currents and voltages are always expressed in 'root-mean-square' (r.m.s.) values which is the value of an a.c. current that does the same amount of work as a given value of d.c. current. The r.m.s. value for a sinusoidal current (and voltage, as voltage and current are proportional) is 0.707 times the peak or maximum value.
You can work this out yourself. For a sinusoidal waveform the rms value is 0.707 times the peak value. As you quote a peak-to-peak value, this must be halved, first. Incidentally, the symbol for volt is 'V', not 'v'.
AC voltage, like the voltage in your house, is typically referred to as 120vAC. This means the voltage swings 120V positive and 120V negative 60 times per second (60 Hz) 360 degrees total (sine wave). Current and voltage go hand-n-hand so the current alternates with the voltage. the RMS value is what we experience at the output (160vAC is actually sent to the circuits)