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Resistive Load and Battery Life

Russell368

Dec 12, 2014
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Hello all,

This is my first post and I am definitely new to electrical theory and electronics, tho I do have a foggy grasp of the basics.

My question is about a simple circuit with a battery and an resistive load, such as a heating coil. I understand that the instantaneous power is equal to the volts * amps, and that the amps will decrease as the resistance increases. So, does that lead me to the conclusion that increasing resistive load in a circuit will lower the average power over time in the circuit and also decrease the battery discharge rate?

This seems counter intuitive to me, what am I missing?

Thanks in advance.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Just goes to show that intuition will often fail you. "Increasing the resistive load in a circuit" could have two interpretations. Increasing the load could mean you are demanding more power from the battery, similar to adding the weight a weight-lifter lifts, increasing their load. This means you are decreasing the electrical resistance of the load. Or your statement could mean you are increasing the load resistance, which means less average power over time and a decrease in the battery discharge rate.

Best to replace intuition with Ohm's law and Kirchoff's Law. For power, use P = E x I = E² / R = I² x R for resistive loads.
 
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Russell368

Dec 12, 2014
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Thanks. Yes, that is a distinction that matters. If I increase the wattage of a light bulb, given the bulb is my resistive load and the higher wattage filament has greater resistance, in that case the increased power demand offsets the decreased amperage due to the higher resistance? And the result is the battery my bulb is hooked to discharges quicker even tho I have increased the resistance in the circuit?

The other scenario is that you've increased the resistance of the load without increasing the load, or calling for more work to be done by the load, in this case the power and discharge rate stay the same. Such as putting a resistor inline to a motor and the motor receives proportionately less power.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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For a given rated operating voltage, a higher wattage light bulb will have less resistance than a lower wattage light bulb. The higher wattage light bulb will also exhibit increased amperage, due to its lower resistance, than the lower wattage light bulb. Both situations, lower resistance and higher current at the same operating voltage, require more power for the higher wattage light bulb versus less power required for the lower wattage light bulb.

If I increase the wattage of a light bulb, given the bulb is my resistive load and the higher wattage filament has greater resistance ...
The higher wattage filament has less resistance than the lower wattage filament.

And the result is the battery my bulb is hooked to discharges quicker even tho I have increased the resistance in the circuit?
The batter discharges quicker because you have increased the amount of power delivered by the battery to the higher wattage bulb. The resistance in the circuit has not increased, it has decreased.

The other scenario is that you've increased the resistance of the load without increasing the load, or calling for more work to be done by the load, in this case the power and discharge rate stay the same. Such as putting a resistor inline to a motor and the motor receives proportionately less power.
If you increase the resistance of the load then you are in fact changing the load. In the case of increasing resistance, the load is decreasing and less power is provided to the load.
 

Russell368

Dec 12, 2014
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Ok, that makes sense to me now that a higher wattage bulb has less resistance, as watts are power (time rate of work), so less resistance means more power and consumes more energy.

In relation to heat energy from a filament this sort of bugged me, as I thought more resistance to electron flow generated the heat, but now I see the heat comes from the amperage (thats where the energy goes).

Thanks again for helping me dispel my misconceptions, you've been very patient with this amateur.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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... In relation to heat energy from a filament this sort of bugged me, as I thought more resistance to electron flow generated the heat, but now I see the heat comes from the amperage (thats where the energy goes).
No, the heat comes from the amperage flowing through a resistance thus dissipating power: P = I² R
You can circulate large currents in superconducting wire coils without any power dissipation because the R term in the equation is zero. Similarly, the largest resistance in the world dissipates no power (the I² term is zero) unless you can place an electrical potential across it that will cause a current to flow in the resistor.

Thanks again for helping me dispel my misconceptions, you've been very patient with this amateur.
Patient? I must be getting old. Thanks for actually listening. And welcome to Electronics Point. There are plenty of people here that will help you learn about electrical theory and electronics. We all started out ignorant of almost everything, but with time and effort and good teachers almost anything can be learned. I hope you stick around for awhile. I think we have an pretty good group of responders.
 

Russell368

Dec 12, 2014
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I think there must be some piece of theory I am missing to fully understand these relationships. Given two circuits both with 12v heating elements of different resistances, respectively with R1 at 1.2Ω and the second circuit R2 at 1.8Ω. I can see the P=120W for R1 and P=80.8W for R2, now, R1 should deplete the battery faster and produce more heat. 1^2R is Joules law and should give me the heat value in joules = watts.

Like you said, amperage does not equate to heat without resistance, but it seems that heat generated increases as Ω decreases until 0, at which point no heat is generated.
 
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garublador

Oct 14, 2014
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I think there must be some piece of theory I am missing to fully understand these relationships. Given two circuits both with 12v heating elements of different resistances, respectively with R1 at 1.2Ω and the second circuit R2 at 1.8Ω. I can see the P=120W for R1 and P=80.8W for R2, now, R1 should deplete the battery faster and produce more heat. 1^2R is Joules law and should give me the heat value in joules = watts.

Like you said, amperage does not equate to heat without resistance, but it seems that heat generated increases as Ω decreases until 0, at which point no heat is generated.
In this case heat generated increases as resistance decreases because the voltage is constant.

It can all be confusing because you have two equations that you have to keep straight, V = I*R and P = I*V. You can see that when voltage is constant that decreasing resistance will increase current, which will increase power. If current were constant, then increasing resistance would increase voltage which would increase power.
 

BobK

Jan 5, 2010
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What you are missing is that lower resistance means more current, given the same voltage. Since the equation for power has the current squared, the increase in current override the lower resistance.

P = I^2 * R

Drop R by a factor of 2 with the same current, and the power is now

P = I^2 * R / 2.

In other words, the power is cut in half.

But if you drop R by a factor of 2 with the same voltage, I goes up by a factor of 2 and:

P = (I*2)^2 * R / 2

P = I^2 * 4 * R / 2

Or twice the original power.

Bob
 

Russell368

Dec 12, 2014
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Given wasted power (heat) = I^2*R (that the heat is proportional to the square of the current by the resistance), how do I account for the internal resistance of the circuit to find at what element (external) resistance the most power is used to make heat? And what relationship does this have to total wasted power, in that, can poorly matched resistances will generate more heat, what is the trade off and curve therein and how is the extra heat generated by impedance matching accounted for by the math?
 

BobK

Jan 5, 2010
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Matching the external resistance to the batteries internal resistance would indeed result in the most power transferred to the external resistance. But it is not likely that is what you want, since it would result in equal heating of the battery. What you want is most of the heat in the external resistance, which means a much higher resistance than the internal resistance of the battery.

The battery should be sized to achieve the heat required in the heating element without much heat wasted in the battery.

Also note that the internal resistance of the battery will increase significantly as the battery is used up.

Bob
 

BobK

Jan 5, 2010
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Okay, you asked for some math.

Let's do it by example:

You have a battery that is 10V and internal resistance 10Ω.

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For a heater coil with 5Ω you would get:

I = 10 / (5 + 10) = 10 / 15 = 0.667A

Power in the heater is: 0.667^2 * 5 = 2.22W
Power in the battery is 0.667^2 * 10 = 4.44W
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For a heater coil with 10Ω you would get:

I = 10 / (10 + 10) = 0.5A

Power in the heater coil is 0.5^2 * 10 = 2.5W
Power in the battery is the same: 0.5^2 * 1- = 2.5W
---------------------------------------------------------
For a heater coil with 20Ω you would get

I = 10 / (10 + 20) = 0.33A

Power in the heater coil is 0.33^2 * 20 2.22W
Power in the battery is 0.33^2 * 10 = 1.11W

Bob
 

Russell368

Dec 12, 2014
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That's great, thanks! So does the difference in wattage produced at the element v the battery relate to where the most voltage is dropped? i.e., where the most power is wasted as heat?
 
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BobK

Jan 5, 2010
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Yes, with the same current through the batteries internal resistance and the external load, the one that drops more of the voltage (i.e. the higher resistance) will dissipate the most power.

Bob
 
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