# Measuring power consumption

weehian
to measure the amount of power drawn by an appliance, can we simply multiply the voltage (V) of the mains by the current (I)drawn (measured using a digital clamp meter) by the appliance to obtain power (P=I x V). does the power company charge us (rate x Power) based of this value P, regardless of whether the electrical load of the appliance is resistive or inductive?

thanks thanks. ## Answers and Replies

Gold Member
weehian said:
to measure the amount of power drawn by an appliance, can we simply multiply the voltage (V) of the mains by the current (I)drawn (measured using a digital clamp meter) by the appliance to obtain power (P=I x V). does the power company charge us (rate x Power) based of this value P, regardless of whether the electrical load of the appliance is resistive or inductive?

thanks thanks. Bigger factories have an special charge if the electrical load is too inductive. In order to avoid it, they usually employ a set of condensers for balance the power factor ($$cos\phi$$). For smaller consumers like homes, I think there is no charge for having an inductive $$cos\phi$$. But I am not too sure.

Mentor
You are both correct (for the US, anyway).

One minor clarification though: there are two wires going to an appliance (plus the ground), but you only measure the current in one of them (they should be the same). Essentially, one is the input (hot) and the other the output (neutral). You may have known that, weehian, but it wasn't clear in your post.

One added fyi: commercial users are also charged for demand - a separate charge for the biggest power draw (measured in 15 min, 30 min, or 60 min increments) seen that month.

weehian
thanks Clausius2 and Russ,

i understand that in most households, power companies charge users based on the resistive power, which is measured using a watt-hour meter. (pls correct me if i'm wrong)
I do not have a separate watt-hour meter at home to measure power consumption of each appliance.
my question next is, when using a digital clamp meter to measure current (I) through the appliance, and then multiplying the reading by 240Volts(mains voltage in my country), is the value obtained the apparent power or the resistive power?

sorry for the confusion in my previous post. many thanks. Last edited:
Gold Member
weehian said:
when using a digital clamp meter to measure current (I) through the appliance, and then multiplying the reading by 240Volts(mains voltage in my country), is the value obtained the apparent power or the resistive power?
:

It is the apparent power, but both resistive and apparent will be surely similar in your home, because you will have $$cos\phi \sim O(1)$$, although I am not sure too. Maybe this could help you:

i) Apparent complex power: $$S=UIcos\phi+jUIsen\phi=P+Qj$$ where
$$j$$ is the imaginary unit, $$P(Watts)$$ is the resistive (Active) power, $$Q(VAR-Reactive-Volts-Amperes)$$ is the inductive (Reactive) power, and $$S(Volts-Amperes)$$ is the apparent power.

You can check that the modulus of $$S$$ is precisely $$S=UI$$, and that when $$cos\phi \rightarrow 1$$ (ie the load is more resistive), then $$S\rightarrow P$$

weehian
Clausius2 said:
It is the apparent power, but both resistive and apparent will be surely similar in your home, because you will have $$cos\phi \sim O(1)$$, although I am not sure too. Maybe this could help you:

i) Apparent complex power: $$S=UIcos\phi+jUIsen\phi=P+Qj$$ where
$$j$$ is the imaginary unit, $$P(Watts)$$ is the resistive (Active) power, $$Q(VAR-Reactive-Volts-Amperes)$$ is the inductive (Reactive) power, and $$S(Volts-Amperes)$$ is the apparent power.

You can check that the modulus of $$S$$ is precisely $$S=UI$$, and that when $$cos\phi \rightarrow 1$$ (ie the load is more resistive), then $$S\rightarrow P$$

thanks claudius2 for your very detailed explanation. if that's the apparent power that i measured, it will not be reflected accurately in my electricity bill right?
i'm actually calculating how much power my aqurium pump is drawing and how much i'm paying for it. my aquarium pump states a consumption of 130W, but the apparent power that i measured using the reading from a digital clamp meter x voltage (240V) is 200W.
my 1st question is, does tampering with the capacitor to improve power factor bring savings in the electricity bill?
and when it says 130W, am i really paying for only 130W of electricity and not more. thanks, i didn't have a strong physics foundation back in school. Mentor
weehian said:
thanks claudius2 for your very detailed explanation. if that's the apparent power that i measured, it will not be reflected accurately in my electricity bill right?
Yes, essentially you get a freebie (caveat: nothing is free and power companies work a correction into their rates for this). Since the vast majority of the electric usage in a residence is resistive, power companies don't bother with trying to figure out the power factor for homes. So yeah, if you run more electric motors than average, you'll get a few percent free.
1st question is, does tampering with the capacitor to improve power factor bring savings in the electricity bill?
No, it will increase your electric bill, if anything. HERE is some info on this:
Let's take an example. A 3/4 HP electric motor has a power factor of .85. The nameplate current is 10 Amps at 115 Volts, or 1150 Volt Amps.

Apparent power = 1150 Volt Amps
Active power = .85 * 1150 = 977.5 Watts
Reactive Power = sqrt(1150^2 - 977.5^2) = 605 VAR
Now, my EE gets a little thin here too, and I need to brush up on this, since I'm currently measuring the power usage by a bank of elevators...

For one, I'm not sure if there is a "typical" power factor of an electric motor - if you plug in the two powers you have, though, you may be able to calculate the pf of yours.

Gold Member
russ_watters said:
For one, I'm not sure if there is a "typical" power factor of an electric motor - if you plug in the two powers you have, though, you may be able to calculate the pf of yours.

The power factor of our friend is $$cos\phi=130/200=0.65$$. I will say don't worry about being charged 200VA, I think you will be charged 130W because I do think it is what a power meter will measure. I have never seen a reactive power meter in a home.

weehian
russ_watters said:
Yes, essentially you get a freebie (caveat: nothing is free and power companies work a correction into their rates for this). Since the vast majority of the electric usage in a residence is resistive, power companies don't bother with trying to figure out the power factor for homes. So yeah, if you run more electric motors than average, you'll get a few percent free. No, it will increase your electric bill, if anything. HERE is some info on this: Now, my EE gets a little thin here too, and I need to brush up on this, since I'm currently measuring the power usage by a bank of elevators...

For one, I'm not sure if there is a "typical" power factor of an electric motor - if you plug in the two powers you have, though, you may be able to calculate the pf of yours.

thanks russ for the link to the excellent article and explanations. i'm clear about this topic now. weehian
Clausius2 said:
The power factor of our friend is $$cos\phi=130/200=0.65$$. I will say don't worry about being charged 200VA, I think you will be charged 130W because I do think it is what a power meter will measure. I have never seen a reactive power meter in a home.

thanks claudius, i'm very glad to hear that. my total apparent power my aquarium equipment is about 2.5kwh, while the sum of the stated power consumption of each equipment adds up only to 1.5Kwh.

very glad that i'm actually paying for the latter. thanks for all the very prompt replies and thorough explanations.

Antiphon
All of this assumes that you have sinusouds everywhere. If you don't (such
as with a non-linear magnetic load) you need a special power measuring
instrument.

What it does is digitally sample the voltage and current at high frequencies and
perform the numerical claculations of VI and RMS power and all that in a microcomputer.