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Voltage-timeIn a Voltage time graph , at constant current I= 170 amps, voltage=0 to 5,5 and time= 0 to 97 sec and voltage decreases from 5,5 at 0sec to 1,16 at 97 sec, now how can I find the total power consumption wrt time in this case and in a compley system how can we find power losses in the cables or circut please some one help me out. Thank you

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  • \$\begingroup\$ Welcome to EE.SE. Please add the graph into your question and fix the punctuation, show your work so far and where you are stuck. \$\endgroup\$ – Transistor Jul 5 '18 at 11:56
  • \$\begingroup\$ Instantaneous power is the (instantaneous) product of voltage and current. When you ask "how much is the power input to the circuit" and "how can we find power losses in the system", it's like asking if it's raining at your hometown without telling us where is your hometown. If it's more of a "theoretical question", then keep in mind that energy conservation is a thing. \$\endgroup\$ – Vicente Cunha Jul 5 '18 at 12:18
  • \$\begingroup\$ I was trying a resistive heating on a material \$\endgroup\$ – Pru Jul 5 '18 at 12:28
  • \$\begingroup\$ That doesn't show the work you did, only a repetition of the problem, on paper. \$\endgroup\$ – a concerned citizen Jul 5 '18 at 13:09
  • \$\begingroup\$ But how can we do integration of that shaded area? \$\endgroup\$ – Pru Jul 5 '18 at 13:52
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Power = volts x amps.

And, given that the current is constant at 170 amps, you can multiply your voltage graph by 170 to get a power graph (watts) versus time.

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  • \$\begingroup\$ Hi thank you for the answer, can you please help me find the shaded area under the curve please. Thank you \$\endgroup\$ – Pru Jul 6 '18 at 8:25
  • \$\begingroup\$ @Pru "shaded area under the curve" = maths problem and not an EE problem. You need to find the average voltage of your waveform over the 97 seconds (hint). \$\endgroup\$ – Andy aka Jul 6 '18 at 9:50

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