On the variation of the power of a resistor with a change in its resistance

So I've been reading up on electronics to prepare for a college project. At this one website, I read that, as a result of long term exposure to heat, the resistance of a resistor goes up, and since V = IR, the voltage across it goes up as well. Also since P = IV, the power of the resistor goes up, so it eventually gives off too much heat and burns up. There was something I didn't like about this explanation. When the resistance increases, wouldn't the current through the resistor decrease? So who says the voltage drop increases more than the current decreases? (Of course, the voltage drop is merely across the resistor itself, not the whole circuit. For a fixed voltage source, the voltage of the source remains.. fixed.) Besides, using V = IR in the equation P = IV gives us P = I² R. So the power depends upon the square of the current, giving it even more importance. To settle the matter, I decided to get a rule for whether the power increases or decreases.



As in the diagram, we take a fixed voltage source V. Let the resistance of the resistor in question be R₁, and the combined resistance of everything else be R. Let the current that originally flows through the circuit (and through R₁) be I₁, and the current that flows through the circuit after the resistance of the resistor is changed to R₂ be I₂. And likewise, let the original voltage drop across R₁ be V₁, and the new voltage drop be V₂. Additionally, let us assume that R₂ > R₁ (i.e. the resistance of the resistor is increased)



For the whole circuit,
V = I₁ (R₁ + R) -> I₁ = V / (R₁ + R)
V = I₂ (R₂ + R) -> I₂ = V / (R₂ + R)

Now let's evaluate the initial and final power (P₁ and P₂) across the resistor:

P₁ = I₁² R₁ = [(V / (R₁ + R)]² R₁

P₂ = I₂² R₂ = [(V / (R₂ + R)]² R₂

Then, P₂ – P₁ = V² [ R₂/(R₂ + R)² – R₁/(R₁ + R)²]

= V² [R₂/(R₂² + 2R₂R + R²) – R₁/(R₁² + 2R₁R + R²)]

= V² [1/(R₂ + 2R + R²/R₂) – 1/(R₁ + 2R + R²/R₁)]

= V² (R₁ + 2R + R²/R₁ – R₂ – 2R – R²/R₂)
   ----------------------------------------------------------
    {(R₂ + 2R + R²/R₂)(R₁ + 2R + R²/R₁)

= V² {R₁ – R₂ + R²(1/R₁ – 1/R₂)}
    ----------------------------------------------------
    (R₂ + 2R + R²/R₂)(R₁ + 2R + R²/R₁)

Now, V² > 0 and the denominator is a positive number as well (sum and product of positive numbers).

Therefore, for P₂ – P₁ > 0 (i.e. for the power to increase),

R₁ – R₂ + R²(1/R₁ – 1/R₂) > 0

or, R²(R₂ – R₁)/(R₁R₂) > R₂ – R₁ -------------- (1)

Since we assumed R₂ > R₁, (R₂ – R₁) is positive, so we can cancel it from both sides without changing the inequality sign.

R²/(R₁R₂) > 1

R² > R₁R₂

R > √(R₁R₂) ------------ (A)

To consider the alternative case, when the resistance is decreased i.e. R₂ < R₁, so (R₂ – R₁) < 0,

When cancelling (R₂ – R₁) from (1), we change the sign of the inequality:

R²/(R₁R₂) < 1

R < √(R₁R₂) ------------- (B)

What (A) and (B) tell us:

- If the resistance of the resistor is decreased, then the power of the resistor will increase IF the total outside resistance R is greater than the geometric mean of the old and new resistances of the resistor. [R > √(R₁R₂) ] Conversely, if R < √(R₁R₂), then the power of the resistor will decrease.
- On the other hand, if the resistance of the resistor is increased, then the power of the resistor will increase IF R is less than the geometric mean of R₂ and R₁. [R < √(R₁R₂) ]
Conversely, if R > √(R₁R₂), then the power of the resistor will decrease. 

Some implications: - If a resistor is part of a circuit with a much higher overall resistance, then if you increase the resistance of the resistor itself, the power across it will increase, thereby heating it up.

- Alternatively, if you take the only resistor in a circuit and decrease it (with the resistance of the wires etc being very low compared to that of the resistor), the power across the resistor will increase as well. If it is decreased low enough, it could result in a short-circuit, burning up the resistor.