Power loss through resistance?

[rubixx]

OK, well the forum says elementary questions, so should be easy ..

My question is:

If i increase the voltage in my circuit by adding extra batteries in series, and then drop to nominal voltage
with a resistor, will 100% of excess voltage be dissipated by the resistor in form of heat and leave me with the same mAH as when i started?
Given connecting batteries in parallel would be easier in most cases, but in a flashlight tube, for example, when there is 0 tolerance between the cylinder walls and batteries and adding a resistor would be easier.. I cant figure this out!!

 

[alexan_e]

If i increase the voltage in my circuit by adding extra batteries in series, and then drop to nominal voltage
with a resistor, will 100% of excess voltage be dissipated by the resistor in form of heat and leave me with the same mAH as when i started?

The way you put is yes but

a) I don't see the reason to increase the voltage just to heat a resistor without any other benefit

b) the voltage drop across the resistor will vary according to the load current so you can't expect a constant voltage drop

There are constant current driver that can be used (usually with leds), these can actually get advantage of the higher voltage and consume lower battery power and last longer.
These regulators use the switching regulator principle using PWM so the efficiency is high.

Alex

 

[snicolasss]

@rubixx,

If your Flashlight is based on LED it has a constant voltage drop, so the resistor will drop the extra voltage, but be careful, remember Ohm's law:
Current = Voltage/Resistance
Then the extra voltage will automatically rice the current, and it can damage your tube.

good luck.

-Snicolasss

 

[rubixx]

Thanks for the feedback, its still a bit cloudy for me so ill exaggerate the situation..

If i have 1x 10V 1000mAh battery and a light that consumes 1A at 10V it should run for about 1 hour right?
So now I connect 10x 10V 1000mAh batteries in series for 100V(still 1000mAh) and a resistor (90Ω) ? Will the lamp run for longer (><10hs) before the batteries go dead? because it should be 100Wh of capacity in the batts and the lamp draw 10w? Or is resistor sending other 90w to burn the butts of passing flies and only last 1hr the light?

Sorry for so complicated but my head wants to explode...

 

[vinodstanur]

If it is a 1 Ah battery then it will get finished by 1 hour if you take constant current of 1A.
So, even if the batteries are connected series , 1 A flows through each cell in your case. So, all will get finished by 1 hour....

---------- Post added at 21:08 ---------- Previous post was at 21:00 ----------

Yes it is having 100Wh capacity if it is connected in series. But at the same time the resistor takes 90 W and the bulb takes 10 W, a total of 100 W. So, it will get finished by (100 Wh/ 100 W) = 1 hour