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After reading this answer, I got to know that battery lasts about Total charge divided by current draw.

So, lets assume the voltage across 2 terminals of battery is constant. by, $$ V(Constant) = IR$$

If I apply high resistance load in series, then very low current will flow and the battery will be drained even more slowly.

My question is, Is my understanding correct(or I am missing something)?

(I have never made any circuit other than led in series,parallel)I was going to use LDR to turn on and off led lights. Earlier i used to think that high resistance produces heat$$H=I^2Rt$$ and that would consume a lot of energy wasted in heating. so battery will be drained faster. But today i read that above linked post and my concepts changed.

(I don't know whether this question was asked before. because I don't know how do I search for it. I searched battery and resistance but irrevelant results came. so I asked question)

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    \$\begingroup\$ I don't understand you question. Power of a resistive load is \$\frac{V^2}{R}\$. So yeah, the higher is \$R\$ the lower is the power... \$\endgroup\$ – Eugene Sh. Nov 8 '16 at 14:33
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    \$\begingroup\$ When an input is constant use the more suitable formula. with otherwise if I is constant, use I. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 8 '16 at 14:38
  • \$\begingroup\$ My question is, today what I read and then applied ohms law. Is that correct? \$\endgroup\$ – Fennekin Nov 8 '16 at 14:40
  • \$\begingroup\$ because I don't know how do I search for it Uhm, the search bar on the top right ? Search for "resistor dissipation". \$\endgroup\$ – Bimpelrekkie Nov 8 '16 at 14:48
  • \$\begingroup\$ Resistor dissipation. Hmmm. I was searching for battery , resistance \$\endgroup\$ – Fennekin Nov 8 '16 at 15:14
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For purely resistive loads, power dissipated can be calculated with $$ P = IV $$ where I and V can be substituted using Ohm's Law, eg: $$ P = I (IR) = I^2R $$

Hence where your H equation comes from. That just takes into account time as well for a total loss over time. To answer your concern about high resistance equaling high heat, higher resistance means lower current. Because current is the squared term in the power equation, it tends to have more mathematical "influence" on the total power consumption. You'll find that a 10ohm resistor with 10V applied (1 amp, 10 watts!) will have higher power than a 100ohm resistor with 10V applied (100mA, only 1 watt). So in fact, a lower resistance will result in higher current and brighter LEDs at the cost of faster battery drain and more heat dissipation via resistors.

The answer you linked has way more on this, but for completeness and to summarize: Battery capacities are measured in millamp hours(mAh), which is simply how many milliamps it can put out for 1 hour. A 500mAh battery can output 500mA for approximately 1 hour. The relation is linear too, so that same battery could put out 250mA for 2 hours or 1000mA for half an hour.

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  • \$\begingroup\$ So current is squared that is the main reason for heat generated. \$\endgroup\$ – Fennekin Nov 8 '16 at 15:12
  • \$\begingroup\$ @Fennekin The "main reason for heat generated" is that the resistor uses power. \$\endgroup\$ – user253751 Nov 9 '16 at 1:55
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A battery has a nominal charge capacity \$Q_N\$ in the order of 1 to 100 Ah for usual consumer variants (AA, mobile, car battery ...). The battery also has a nominal voltage \$U_N\$ related to its chemistry and architecture.

An oversimplified model of a battery gives a constant output voltage until its charge is depleted. If you apply a constant load to this (e.g. resistors and LEDs, no switching), then the time you can run the load from this battery could be calculated by
\$t = \frac{Q_n}{I} = Q_n\cdot\frac{R_{total}}{U_N}\$, where \$I\$ is the constant discharge current.

As can be seen, increasing the Resistance \$R_{total}\$ increases the battery runtime. But the Question is whether the circuit is still able to fulfill its purpose. A LED which receives less current due to a higher series resistor will emit less light, possibly making it useless for room lighting or as torch light.

Real batteries however are far more complicated and have to be modeled at least with a series resistor. There are two effects in a battery which will actually increase its useable charge \$Q_u\$ over its nominal charge capacity when you apply smaller discharge currents (~ higher load resistance) :

  • The Peukert Effect describes that one can get more charge out of a battery if discharged with a low (constant) current.
  • The Recovery Effect says that in periods of low/no discharge currents the reduced "useable" charge due to high current loads gets partially replenished.

The reasons for both are the chemical processes in the battery. Therefore battery runtime is affected by a lot more factors than Ohm's Law alone. These effects do not create charge magically out of nowhere, but rather show that the discharging process is more effective in the given circumstances of lower currents.

Parts of this answer are excerpts from an related older answer of mine.

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While your formula to calculate the heat generated by a resistor is correct, it is not useful in this case. This is because if you change the resistance, but keep the battery at the same voltage, the current will change as well as the resistance, so you can't deduce that more heat is generated if you increase the resistor that loads the battery.

Because you have a fixed input voltage, you should use a formula expressing the heat in terms of \$U\$, \$R\$ and \$t\$:

\$H = \frac{U^2}{R}t\$

This formula shows clearly that if voltage and time stay the same, less heat is generated for a higher resistance. You get my formula from your forma if you solve Ohm's law for \$I\$ and substitute the \$I\$ in your formula with the result.

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  • \$\begingroup\$ Thats what eugene commented \$\endgroup\$ – Fennekin Nov 9 '16 at 1:05
  • \$\begingroup\$ It also is the answer to your question. If @eugene posts it as answer in a comparable level of detail, I am happy to upvote it and delete my answer. While the other answers are quite informative, I consider them to miss the main point of the question. Eugene's comment is in fact spot on, but misses an explanation why you should choose the other formula. Tony's comment addresses that shortcoming, but still does not explain the relationship between the formula in the question and the actually useful formula. \$\endgroup\$ – Michael Karcher Nov 9 '16 at 5:43
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Theoretical answer is YES, But it depend on where you use it. If it is for HIFI system, The sound output will goes bad. Higher power source resistance will lower the fast response ability.

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