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I am a tenth grade student.

Actually I have two questions.

  1. I read that internal resistance of a cell decreases if the temperature of the electrolyte increases. But isn't it contradicting what is said before: "resistance increases with increase in temperature".

I think if temperature will increase then the molecules and the ions of the electrolyte will move more randomly, as a result of which the charges inside the battery while moving towards the electrodes will suffer more collisions resulting in increase in internal resistance.

My next question is:-

  1. in our book it was written "when current flows in a circuit heat is generated", and to calculate the heat it was given

i)heat generated is directly proportional to square of current

ii)heat generated is directly proportional to resistance of the circuit.

iii)heat generated is directly proportional to time.

Combining the equations (i),(ii)&(iii) along with proportionality constant we get:-

$$H=I^2Rt$$

But if we modify the equation, that is suppose we take $R=\dfrac VI$, where $V$ is the voltage and $I$ the current in the circuit and substitute it, then we will get

$$H=\frac{I^2Vt}I$$

and here it says H is directly proportional to $I^2$ as well as H is inversely proportional to $I$. How can this be?

Even if we cancel out the two $I$ from the numerator and the denominator it will state H is directly proportional to $I$.

Previously it was mentioned that $H$ is directly proportional to $I^2$ and now it mentions it is directly proportional to $I$

and here stands the contradiction. If we replace the value of $I^2$ in terms of V and $R$ we find a similar contradiction occurring.

I hope there is a possible explanation to this and would be highly obliged if somebody could kindly help me out.

Thank you!!

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  • $\begingroup$ Hi and welcome to physics SE. Please, use laTex notation for formulae. It's about writing them in between of dollar symbols, and laTex commands inside. See here: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – FGSUZ
    Aug 11 '20 at 21:50
  • $\begingroup$ That is to say, if you want to write $P = I^2 R$, then you'd write $ P = I^2 R $ $\endgroup$
    – David
    Aug 11 '20 at 23:02
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For your first question, let's first talk about the case of a wire. A chunk of metal is a crystal lattice of regularly spaced atoms. When the atoms are perfectly regularly spaced, the momentum of the electrons can be transferred throughout the system without dispersion. As we increase the temperature, the atoms start to bump around, causing the electrons to exchange momentum with the atoms. Collectively, this momentum exchange results in energy dissipation which we measure as resistance.

An electrolyte is different: charge is transferred through diffusion. Diffusion is a process that occurs if, for example, you mix milk with coffee: the milk particles and the coffee particles diffuse into one another because microscopic interactions between their molecules cause them to mix. As we increase temperature, we increase the rate of these interactions allowing them to mix faster. For an electrolyte, the charge carrying particle effectively takes a random walk with a bias caused by the applied voltage. This process is just like diffusion, in that the amount of "steps taken" per unit time increases as the temperature increases, thus increasing the conductivity of the electrolyte.

For your second question, you must keep in mind what you are holding constant in your circuit. For example, if we change $I$ and $R$, then $V$ is determined for us because $V = I R$! In other words, of $V$, $I$, and $R$, only two are independent. So, when you compute the power dissipated by your circuit, you often want to write the result in terms of the quantities you are manipulating, while eliminating the one you aren't. As David White points out, three common choices are $P = I V$ if $R$ is constant, $P = I^2 R$ if $V$ is constant, and $P = V^2/R$ if $I$ is constant. These are all equivalent, but which one is useful depends on what you know about your circuit.

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For question 1, note that the velocity of the ionic species in the battery's electrolyte will increase as the temperature increases. This will increase ionic mobility, and should decrease the measured internal resistance of the battery as a result. Your comment about resistance increasing with temperature is what one would expect when dealing with a conductor (e.g., a wire), not an electrolyte.

For question 2, there are three equations to calculate electrical power:

  1. $P=IV$

  2. $P=I^2R$

  3. $P=V^2/R$

Power equals work divided by time, so the total heat generation will be

$H=Pt$, where you can use any of the three equations above for the power term, depending on what your known values are.

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I read that internal resistance of a cell decreases if the temperature of the electrolyte increases. But isn't it contradicting what is said before: "resistance increases with increase in temperature".

While it is true that things like copper wires have higher resistance with increase in temperature, resistance doesn't always increase with increase in temperature. For example, a thermistor that decreases its resistance with an increase in temperature, i.e., it has a negative temperature coefficient. Insofar as a battery electrolyte is concerned, it is my understanding that it becomes more conductive as temperature increases because because it increases the movement of ions.

But if we modify the equation, that is suppose we take R=V/I, where V is the voltage and I the current in the circuit and substitute it, then we will get

$H=(I^{2}Vt)/I$ and here it says H is directly proportional to $I^2$ as well as H is inversely proportional to I. How can this be ? Even if we cancel out the two I from the numerator and the denominator it will state H is directly proportional to I.

$H$ can only be proportional to $I$ if $V$ is a constant. But for a resistor $V$ is not independent of $I$. They are related by Ohms law. So you need to take that final step $I^{2}/I = I$ and you get

$H=(VI)t$

Which is equivalent to

$H=I^{2}Rt$ and

$H=V^{2}t/R$

Hope this helps.

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