# Power and bulbs I have noticed the use of two different formulas i.e P=V²/R (in the 3rd line) and P=I²R (5th line from below). Since the bulbs are in series, how can the resistance be inversely proportional to power as shown? Is anything wrong with my book or am I missing some fundamental information?

• Have you done the math? Nov 6 '17 at 14:22
• Both formulas are sound and a combination of Ohm's Law V = I * R with the equation for power P = V * I Fill in one into the other and you will get the same equations. Nov 6 '17 at 14:24

You only need to remember two formulas. The rest can be worked out.

$$V = IR \tag 1$$ $$P = VI \tag 2$$

From (1) we can say that $I = \frac {V}{R}$ so popping that into (2) we get $$P = V \frac {V}{R} = \frac {V^2}{R} \tag 3$$

Alternately we pop (1) straight into (2) and say $$P = VI = IR \cdot I = I^2R \tag 4$$

That's the background sorted.

Now, for a given current $P = I^2R$ tells us that $P \propto R$.

Finally - and this may be the confusing part - note that the lamps are designed to be run in parallel in normal operation so that they run at the same voltage. In that configuration the lamp with the lower resistance will pass more current and dissipate more power (have a higher wattage).

In this application, because you are running in series they run at the same current and the lamp with the higher resistance will dissipate the higher power.

• which formula for power should i use in case of a single resistor rather than a combination if resistors? Nov 6 '17 at 15:59
• The one that contains the unknown and two known variables, of course! Nov 6 '17 at 16:14

Both the bulbs have different Power rating, but they are assumed to have the same voltage rating in this question. So it's obvious that the bulb with less power have more resistance, from the equation P = V^2/R. Now in series connection, the current is same in both the bulbs. So you can apply P= I^2*R formula now. Thus the 60 W bulb which has more resistance, will develop more power and hence glow brighter.

• Very nice induction. The secret being the current is absolutely the same in both bulbs no matter what voltage conditions exist. If the current is the same then power relates to directly to R. So you only need to know P=I^2*R. Nov 6 '17 at 16:43

Simple answer: We know from Ohm's Law and Thevenin voltage drop in a loop, that the most voltage is dropped on the highest resistance and the lower P rated bulb has a higher R value so it gets brighter at first.

It is assumed you know all the variant's of Ohm's Law and any or all may be used. What you were not told, but is not essential for the basic understanding, is that the bulb current is not constant and due to positive R vs T or PTC characteristic so the smaller bulb wattage which has higher R value warms up faster and raises it's R so it gets near full brightness in steady state < 1 second. The simple answer is the bulb with the highest R or lowest W value is the brightest.

Real Life situation: simulate this circuit – Schematic created using CircuitLab

Tungsten wire in bulbs are also PTC resistors. R_cold (25'C=298'K) = 10% of R_hot (3500'K)

• This means a 100W bulb uses up to 1kW during the 1st cycle turning on ! (depends on start phase , contact bounce etc.)

Thus R1~14Ω , R2~24Ω when off. ( room temp 298'K)

Pedantic note: ... 25'C may be warm for you , if you prefer 21'C with A/C but it is the "de facto" std temp for electronic specs. in datasheets

• So when turned on V(R2)=24/(24+14.4)*Vac = 62.5% Vac
• or V(R2)=75Vac, V(R1)=(120-75)=45Vac

• I init. = 120V/(14.4+24)=3.1A approx avg for 1st cycle period

• which is shorter than filament thermal time constant.

Since current is shared, P=VI , during inital power on cycle, we estimate as follows;

• (roughly 30% accurate and not with 4 sig figs as shown in calculator result below)

• P1=139.5W ( rated 100W)
• P2=232.5W ( rated 60W)

Perhaps counter-intuitive at first, but the bulb rated for the lower power starts with a higher power

• which means it gets hotter faster
• and since the resistance increases with temperature rapidly x10, R2 will almost reach full brightness >200 Ω
• while R1 which started at 14 Ω gets less and less voltage drop as R2 heats up faster
• and since the current drops thru both, R1 just pops then shuts down while R2 heats up slower than nomal
• since R1 cold is only 5% of R2 when hot and the higher R drops the most voltage and hence power
• the normal current for 60W=120V*0.5A is almost achieved since at steady state
• P1 would then only be lets say 10%Vac or 12V* *0.5A=6W so hot but not 3500'K and unlikely visible heat.

Concepts to understand: PTC, Ohm's Law

Intuitive Rule. Highest series impedance gets the most voltage drop.

We call this series PTC or "positive temperature characteristic" characteristic in parts named as this in catalogs, as "over-current protection devices". ( They are not meant to operate at high T forever (years), just for thermal protection of devices from short circuits.)

PTC's come in radial ceramic or SMD format, generally operate with polysilicon material and operate around 80'C with highly non-linear R near this T unlike tungsten which is more linear with T ('K) from 300 to 3000'K thus 10x the R value. ( roughly)

The key that others have implied but not explicitly stated is that there are two different power values for each bulb.

1. The rated power of the bulb, that is the power the bulb will draw when operated under normal operating conditions.
2. The power the bulb will draw when connected in the circuit given in the question.

When someone says "a 60W bulb" what they mean is "a bulb that will draw 60W when supplied with it's rated voltage. If not otherwise specified the rated voltage will be the normal mains voltage for wherever you live.

The first part of the answer is talking about deriving the resistance of the bulb from the rated power (which, as tony points out is a very crude approximation). The two bulbs are assumed to have the same rated voltage. So $P=\frac{V^2}{R}$ is the equation used.

The second part of the answer is talking about the behaviour of the bulbs when placed in the series circuit. In the series circuit the current though the two bulbs is the same. So $P=I^2R$ is the equation used.