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When I connect a cell with 100 Ω resistor. voltage across circuit is 1.4 V and current is 14 mA. That means total power of cell is 1.41 x 0.014 = around 20 mW. Right?

Now if I connect two cells in series with same 100 Ω resister the volts across are 2.83 V and current is 27.6 mA

If we multiply them we get around 78 mW.

What I don't understand is if one cell has around 20 mW of power how come two cells have 78 mW instead of double of 20 mW = 40 mW maybe?

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  • \$\begingroup\$ Did you measure the voltage and the current at the same time? If these were two separate measurements with the same instrument, the resistance of the current meter itself affects the measurements. The battery cell itself also has an internal resistance due to the way it is constructed, this will also affect the measurement. \$\endgroup\$ – MarkU Apr 13 '20 at 21:05
  • \$\begingroup\$ upvote for using the correct term cell instead of battery ... you increased voltage and current, not just one ... double the resistance when two cells are used and run your calculations again \$\endgroup\$ – jsotola Apr 13 '20 at 21:07
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    \$\begingroup\$ Because the power for a constant resistance increases as the square of the voltage. $$P = V \times I = V \times \frac{V}{R} = \frac {V^2}{R}$$ \$\endgroup\$ – G36 Apr 13 '20 at 21:09
  • \$\begingroup\$ @MarkU yes i measure the same time. \$\endgroup\$ – Asim Apr 13 '20 at 21:10
  • \$\begingroup\$ @G35 would explain further. U mean if a cell has 1w power than 2 cells would make 4w ? \$\endgroup\$ – Asim Apr 13 '20 at 21:12
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This might help : \$P=\dfrac{U^2}{R}\$

\$P_1=\dfrac{1.41^2}{100} \approx 20\,\text{mW}\$

\$P_2=\dfrac{2.83^2}{100}\approx 80\,\text{mW}\$

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  • \$\begingroup\$ Would u explain further. What is U² and why its squared. I thought power is simply equal to volt x amps \$\endgroup\$ – Asim Apr 13 '20 at 21:17
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    \$\begingroup\$ U is voltage, yes it is volt x amps, you can replace the amps with I=U/R then you get volt x amps = volts^2 / R. It's all about Ohm's law. \$\endgroup\$ – Marko Buršič Apr 13 '20 at 21:23
  • \$\begingroup\$ U is more of a nomenclature to "electric potential", which is practically synonymous to voltage. \$\endgroup\$ – user103380 Apr 13 '20 at 21:30
  • \$\begingroup\$ @KingDuken It depends where do you live in where you went to school. For example: de.wikipedia.org/wiki/Elektrische_Spannung \$\endgroup\$ – Marko Buršič Apr 13 '20 at 21:33
  • \$\begingroup\$ @KingDuken You see, U is voltage and V is potential. In your school it's V is voltage U is potential. But no matter if I say I=U/R or I=V/R I think it should be clear what I meant. \$\endgroup\$ – Marko Buršič Apr 13 '20 at 21:40
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In this case, the power delivered to the resistor is not limited by the power available from the cell. The power is limited by two factors:

  1. the voltage provided by the cell
  2. the resistance of the load

If you put two cells in series then you double the voltage. If the resistance stays the same then Ohm's Law tells us that the current must also double. So, if the voltage doubles and the current doubles, power goes up by a factor of 4.

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  • \$\begingroup\$ Thanks. But what is the method to measure the max power of the cell? \$\endgroup\$ – Asim Apr 13 '20 at 21:19
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    \$\begingroup\$ To measure the maximum power of the cell, lower the resistance until the cell's output voltage halves. Don't do it for long though, because an equal amount of power will be dissipated inside the cell. en.wikipedia.org/wiki/Maximum_power_transfer_theorem \$\endgroup\$ – Bruce Abbott Apr 13 '20 at 21:24
  • \$\begingroup\$ You would need to determine the equivalent series resistance of the cell. You could approach this by using small load resistor values such that the cell voltage starts falling. At the point where the cell voltage is one-half of its unloaded voltage, you are getting the maximum power from the cell...if you haven't ruined it. \$\endgroup\$ – Elliot Alderson Apr 13 '20 at 21:24
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We know that V=IR (ohms law) . when you added 2 cells in series then the voltage applied is doubled but the load connected(resistance) remained same. So, the current also got doubled. Now, power = V*I . So, as now both current and voltage are doubled one would expect power to get four times. This was what you obtained experimentally.

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Extension/demonstration to @Marko Buršič answer:

[1] \$V={R}*{I}\$

[2] \$P={V}*{I}\$

Calculating \${I}\$ from [1]:

[3] \$I=\dfrac{V}{R}\$

Then, replacing [3] in [2]:

\$P={V}*\dfrac{V}{R}\$

Resulting:

\$P = \dfrac{V^2}{R} \$

As an addition, it could be interesting to consider the cell internal resistance which could cause a voltage drop in the cell terminals depending on the current it is sourcing

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Applying Ohm's Law, you have rightly calculated that the power dissipated by a 100Ω resistor across a 1.4V cell is 19.6mW and that dissipated by it across a 2.8V cell is 78.4mW.

The power dissipated by a load cannot not be equated to the maximum power that can be delivered by the source across which it is connected. Your confusion stems from that wrong assumption.

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