Getting the energy consumption in joules over a period when having current and voltage

Question

I'm trying to calculate the energy consumption over a period of 1.3 seconds in joules (J).

I have a sample every 100 ms (0.1 second) with the current and voltage (in volts) (in practice, the samples have a much higher granularity, but for the sake of simplicity ;)).

Time (s)	Current (mA)	Voltage (V)
0.0	30200	3.55
0.1	30000	3.6
0.2	30200	3.65
0.3	30100	3.55
0.4	30000	3.6
0.5	30100	3.55
0.6	30200	3.6
0.7	30100	3.65
0.8	30000	3.55
0.9	30000	3.6
1.0	30200	3.6
1.2	30000	3.55
1.3	30100	3.6

How can I calculate the consumed energy consumption in joules over this period of 1.3 seconds?

My idea was to:

1. Convert mA to ampere by diving each mA value by 1000: e.g., 30.2, 30, 30.2, 30.1, 30, etc.
2. Taking the mean of the current and voltage, 30.09 A and 3.59 V respectively..
3. Multiplying these to get the power: 30.09 A * 3.59 V = ~108 W
4. Since watt is also joules per second we multiply it by 1.3 to get the total consumed energy of this period in joules: 1.3 s * 108 W = 140.4 J.

Is this correct? Does it make sense? Are there better, more accurate ways to measure the energy consumption in joules over a period?

Answer 1

• Convert \$\text{m}A\$ to \$A\$.
• Multiply \$A\$ and \$V\$ for each time period to get \$W\$.
• Multiply \$ W\$ by the length of the time period (\$0.1\text{ s}\$) to get \$J\$.
• Add up all the \$J\$. Simple numerical integration.

Answer 2

Expanding @Elliot's answer to a graphical illustration:

Since you know the values of voltage \$V\$ and current \$I\$ at different time instants, you can calculate instantaneous power \$P = V . I\$ for those time instants.

Then, if you plot the graph of \$P \$ (in watts) against Time \$t\$ (in seconds):

The different discrete samples are joined by lines with the assumption that \$P\$ varies linearly with Time \$t\$. Now, you can estimate the energy consumption \$E\$ in \$\text{joules}\$ as the numerical integration of \$P\$ over \$t\$; or the area of the power plot under the time interval: \$(0.0\text{s} \text{ to } 1.2\text{s})\$. This area is simply the sum of areas of trapezoids--formed by triangle and rectangle--within every time interval of \$0.1\text{s}\$.

This is one way of estimation and another way of estimation is by assuming that the measured \$P\$ remained constant between the time samples. Then, you can plot the graph like step changes:

In this case the math for energy consumption \$E\$ becomes fairly easy, as \$E=\Sigma(P \times 0.1) \text{ joules}\$.

Again, this is same as the area of power plot under the time interval: \$(0.0\text{s} \text{ to } 1.2\text{s})\$; i.e., the sum of areas of rectangles within every time interval of \$0.1\text{s}\$.

Answer 3

What you propose is one way to estimate how much energy was consumed.

A better method is to calculate how many joules were consumed in the time between each pair of samples, assuming that both voltage and current remained constant during that time. Then, just add up the total energy use over time. This is called a rectangular integration.

You can do better still with a trapezoidal integration. Assume that the power consumed changes linearly between sample points. Calculate the energy consumed as the time between data points multiplied by the average power (the average of the power at the beginning of a sample time and the end of a sample time). Again, just add up the energy for each sample time.

These are all just estimates, of course.

Answer 4

How you do it should depend on why you think the readings are varying.

If you think they vary because of reading errors and noise while the device takes a constant current and is fed with a constant voltage, then the way you describe using means will be accurate.

If you think that the device is actually taking a varying current, then it would be better to estimate the energy consumption in each short period, and sum them.

With this data set, the difference is going to be very small. If the readings varied by a larger fraction of the mean, then the results would be more divergent.

Answer 5

You are describing integration, similar to the derivation of Newton and Gauss. For example, in the first 0.1 s, EMF rose from 3.55 to 3.60 V. One can assume that the change is linear over a small range, so the average EMF over the first second would be half-way between, or 3.575 V (well, I've gone beyond the two significant figure data, but I'll round later. My bad.).

In a similar manner, current fell from 30.20 A to 30.00 A over that time, averaging at 30.10 A.

The total energy usage, over the first 0.1 s would be approximately 3.575 V * 30.10 A, 107.61 V*A, or J.

Repeat 12 times and sum the answer. This is simply a sum.

A more accurate answer might be obtained from curve fitting, thereby obtaining a numerical function to describe the change in voltage and in current with time, and solving the definite integral for that function.