# AC Power source

#### crutschow

Joined Mar 14, 2008
30,419
1,i want to manually calculate the current many i know how do i get 0.01A (if given 5.4V and 60Hz)
You need to know the impedance of the load for a 60Hz sinewave.
2.Difference b/w Vrms and V
V the designation for volts.
Vrms indicates the RMS value of the voltage (which is √2 * Vpk of a sinusoidal where Vpk is the peak voltage).
Vdc would indicate the voltage is DC.

#### uppi_777

Joined Feb 26, 2018
42
You need to know the impedance of the load for a 60Hz sinewave.

May i know how to know the impedance of the load for 60Hz sinewave
i checked data sheet but was unable to locate it

#### Alec_t

Joined Sep 17, 2013
13,039
Whatever you're driving with the AC supply is the load, so the load's impedance won't be in the supply's datasheet.

#### MisterBill2

Joined Jan 23, 2018
12,678
If that is an AC supply that you are using, then that displayed current, 0.01 A is a measured value, not a calculated value. I am not aware of power supplies that calculate current. They may calculate watts and even VAR and power factor, but current and voltage are measured values.

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#### BobTPH

Joined Jun 5, 2013
5,453
What are you powering with it?

#### MisterBill2

Joined Jan 23, 2018
12,678
The display image shown in post #1 appears to be on a regulated adjustable AC power supply with internal metering. It is evidently supplying some sort of load.
Normally sine wave AC voltages are assumed to be expressed in RMS volts, because that is most convenient for calculating power, and also the relationship to peak voltage is well understood.
But given only the voltage and frequency with no hint at the load, I see no way to calculate the current.

So it will be useful to get an explanation of what the TS is actually seeking to discover.

#### MisterBill2

Joined Jan 23, 2018
12,678
Here is the data sheet of my AC Power source
View attachment 274232
This is a nice laboratory type of AC supply. What is missing from the description is any mention of measurement accuracy or resolution. And I disagree! The current is measured, it is not calculated. To calculate current one needs both voltage and resistance. The calculation is very simple, but understand I (current) = E (voltage)/R(resistance in ohms).