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Internal Resistance

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codan

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Hi Everybody,

I have attempted to calculate the internal resistance of a carbon cell given in a question.
I am not quite sure if i have it correct but it seems to work out, i have had trouble with this one.
Could you check my answer & see if i am on the right track?

The question is:

Determine the Internal Resistance of a carbon cell if the open circuit voltage is 1.7v & when a 10ohm resistor is connected across the battery terminals the voltage drops to 1.26v.

My Answer is:

I = V/R = 1.26v/10Ω so I =.126A

r = E - V /I = (1.7v - 1.26v) /.126A =3.49ohms

Then using to verify things:

E= V+Ir = 1.26v + (.126A*3.49Ω )=1.7v

Thank You
 
Looks OK to me.

JimB
 
Just try to do it quickly within a few seconds. There are actually several chemical effects going on. Some are short term, some are longer term. Heavy continuous load will deplete available source of needed chemical reactions locally near the electrodes dropping the voltage more and there is a time delay equilibrium that is established. When load is removed the longer term equilibrium will replenish and some of the appearent capacity will recover. The longer term effects raise the total battery Rs over the short term current pulsed Rs.

A more complete model of a battery is a number of R-C time constants in series. First order is just the contact and surface resistance which is measured on a quick pulsed load. Higher order chemical effects have longer time constants. Different battery chemistries have different longer term time constants. NiCad's are pretty short, Zinc-Air batteries are very long.

You may have noticed this effect on a hard starting auto. A long cranking will eventually run down the battery (and raise the Rs). Allow the battery to sit for a minute or two and it will again turn over the engine starter.

You may think of it as blotting a paper towel against a sponge saturated with water. Tapping the surface of the sponge with the paper towel will deplete the surface of the sponge of water. Leave it alone for a few seconds and the interior saturated with water will migrate out to sponge surface, replenishing it. Eventually all the available water within the sponge will be blotted out, exhausting the water charge within the sponge.
 
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It's most likely a homework question and for the purposes of the course he's studying, the chemical reactions in the battery aren't important.

The way of measuring and calculating the internal impedance is correct, but it's true that it will increase as the cell discharges.
 
It is good that you have verified your answer. I always do this as it saves any simple mistakes. That is the good thing about V=IR, R=V/I, P=I^2R, P=VI... they are all related and you can always work backwards using the result you have just worked out to check the sums
 
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