Light bulb brightness increases with power, $P$.

So why doesn't increasing $R$ increase $P$ and hence increase brightness as $P=I^2\cdot R$ due to $P=I\cdot V$ and $V=I\cdot R$?

I read increasing $R$ decreases brightness.

  • 1
    $\begingroup$ You also have$ P=v^2/r $ where v is voltage so increasing r decreases power and thus the brightness $\endgroup$ Commented Feb 6, 2017 at 7:20
  • $\begingroup$ Thanks for the answer. But how would I know if the $R$ in $P=V^2/R$ or $P=I^2*R$ is dominant? I.e. when $R$ increases, $P=V^2/R$ decreases (decreased brightness), but $P=I^2*R$ increases (increased brightness) - giving contradictory answers. How do I know which to follow? $\endgroup$
    – K-Feldspar
    Commented Feb 6, 2017 at 7:23
  • 2
    $\begingroup$ No if resistance increases current decreases right(in series) so both P,I,R are parameters . In second one only p,r are parameters while Dc voltage is a constant $\endgroup$ Commented Feb 6, 2017 at 7:41

4 Answers 4


Power sources can work in two modes : control voltage (CV) or control current (CC).

In CV mode, the voltage is imposed, and the output current is adjusted depending on the load. This is for instance the case at home, where the electrical plugs deliver 100V (or 220V depending of country), regardless of what is plugged in it. In this case, $P=V^2/R$ is the relevant expression as $V$ is a known value.

In CC mode, the current is imposed and the corresponding voltage is adjusted to match the load. While this mode is less familiar in domestic usages, it is often used in electrical engineering - it insures for instance a constant magnetic field in a coil, regardless heat effects which could alter the resistance of the circuit. In this case, $P=RI^2$ is the relevant expression. If you increase the resistance, the power supply will increase the voltage to maintain constant intensity, and the dissipated power is logically higher.

  • $\begingroup$ The 'C' meant 'constant' (although 'controlled' works too). $\endgroup$
    – amI
    Commented Sep 29, 2018 at 16:22

You can use any of these formulas to calculate $P$:

$$P = I^2 \cdot R$$ $$P = \frac{V^2}{R}$$

They are both correct and will give same result. You can not tell which one is "dominant".

But to use these formulas you need to know not only $R$ but also $I$ or $V$. And to analyze these formulas you need to know how $I$ or $V$ change when you change $R$.

In case you connected the bulb to a power supply with produces constant voltage $V$ it's easier to use the second formula. You can use the first one either, but you should remember that when $R$ increases the $I$ changes as well. The result would be the same: $P$ decreases.

If you connect the bulb to a power supply which produces constant current $I$ then both formulas would tell you that $P$ increases when $R$ increases.


The two equations of relevance are

${\rm power} (P) = \,{\rm voltage} \,(V) \times {\rm current} \,(I)$

${\rm resistance}\, (R) = \dfrac{{\rm voltage}\,(V)}{{\rm current}\,(I)} $

From those two equations you can get $P = I^2R$ and $P = \dfrac {V^2}{R}$

Suppose that it is assumed that the resistance of the light bulb does not vary with the voltage across it / the current through it.

In your room you have a light stand with a $240 \, \rm V,\;60\, \rm W$ light bulb in it and you want to replace it with a brighter $240 \, \rm V,\;100\, \rm W$ light bulb.
Using $R = \dfrac {V^2}{P}$ he working resistance of the $60\, \rm W$ bulb is $694\, \Omega$ and that of the $100\, \rm W$ bulb is $576\, \Omega$.
So decreasing the resistance increases the brightness.

The problem with using $P=I^2R$ is that you might get the impression that because the resistance $R$ goes down the power also decreases but in doing that you have assumed that the current $I$ stays constant.
The current is not constant but actually increases by the same fractional amount as the resistance decreases.
But that is not all because in the equation $P=I^2R$ the current is squared and so the fractional increase in the current squared $I^2$ is double the fractional decrease in the resistance $R$.
So overall the power dissipated increase as the current decreases which leads to increased brightness.


$$V = IR$$

$$P = IV = I^2R = \frac{V^2}{R}$$

It is clear from these equations that if $R$ changes at least one of $I$ or $V$ must also change.

So to answer the question of what effect a change in $R$ will have on $P$ we need more information. Specifcially we need to know about the power source that is feeding the load.

If your power source has a constant output voltage then $P$ will be inversely proportional to $R$.

If your power source has a constant output current then P will be proportional to R.

In reality real power sources are neither truely constant voltage or constant current. Lets say our power source is in fact a battery, we model our battery as a voltage source in series with some internal resistance. The battery has an open circuit voltage $V_\mathrm{OC}$ and an internal resistance of $R_\mathrm{BAT}$. Our load has a resistance $R_\mathrm{LOAD}$ and we want to calculate the powered $P_\mathrm{LOAD}$ delivered to it.

$$V_\mathrm{OC} = I(R_\mathrm{BAT}+R_\mathrm{LOAD})$$

$$P_\mathrm{LOAD} = I^2R_\mathrm{LOAD}$$

$$ I = \frac{V_\mathrm{OC}}{R_\mathrm{BAT}+R_\mathrm{LOAD}}$$

$$P_\mathrm{LOAD} = \left(\frac{V_\mathrm{OC}}{R_\mathrm{BAT}+R_\mathrm{LOAD}}\right)^2R_\mathrm{LOAD} = V_\mathrm{OC}^2\frac{R_\mathrm{LOAD}}{(R_\mathrm{BAT}+R_\mathrm{LOAD})^2}$$

The interesting part of this equation is the last part. We find that the power delivered to the load is largest when the resistance of the load is equal to the internal resistance of the battery.


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