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I want to calculate power loss across a resistive load during a 1 ms interval. The experimental setup is such that the voltage and current through the load varies the whole period of 1 ms. I have 10,000 voltage and current values within this 1 ms period. I am exploring the best way to find the power loss in this case. I have tried multiplying each instant's voltage and current (P= vi), and using trapz function in Matlab, I tried to find out the energy loss value (E=trapz(t,P)). However, I want to learn about the better ways to calculate power loss across the load over the 1 ms period.

Thanks.

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  • \$\begingroup\$ Can you share a graph of your P(t) values? Do you have a reason to think your result is not accurate enough? \$\endgroup\$
    – The Photon
    Commented Aug 22, 2019 at 16:12
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    \$\begingroup\$ multiplying each pair of current.voltage values and summing/integrating the power across the interval is the correct way to do it. The only 'better ways' are to take short cuts when you can definitely assume some things about the setup to avoid calculating them all, for instance constant voltage and current across the interval. \$\endgroup\$
    – Neil_UK
    Commented Aug 22, 2019 at 16:51
  • \$\begingroup\$ If I multiply voltage,current and timestamp at each timestamp (E=vit) and sum them together over a period of 1 ms, it gives me an output of 1010. This is different from the result (0.0029) obtained from the MATLAB using trapz function. I am not sure which result is more meaningful. \$\endgroup\$
    – Allison_81
    Commented Aug 22, 2019 at 16:53
  • \$\begingroup\$ 1010 whats? 0.0029 whats? What are your units? \$\endgroup\$
    – Transistor
    Commented Aug 22, 2019 at 17:28
  • \$\begingroup\$ The unit is joule \$\endgroup\$
    – Allison_81
    Commented Aug 22, 2019 at 18:28

1 Answer 1

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If I multiply voltage,current and timestamp at each timestamp (E=vit) and sum them together over a period of 1 ms, ...

You should be using \$ E = \Sigma vi \Delta t \$ where \$ \Delta t \$ is the sample period in seconds, not the timestamp which will increase with each sample.

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