Why does a 60 W bulb glow brighter than a 100 W bulb when connected in series?

Question

In my physics class I have this problem that shows two lightbulbs, one $60\,\text{W}$ and one $100\,\text{W}$ in series, connected to a $120\,\text{V}$ battery.

The problems are:

Which bulb is brighter? (A: $60\,\text{W}$)

Calculate the power dissipated by the $60\,\text{W}$ bulb. (A: $23.4\,\text{W}$)

Calculate the power dissipated by the $100\,\text{W}$ bulb. (A: $14.1\,\text{W}$)

Why is the power dissipated not simply the wattages of the bulbs? I followed one solution online where you first find $R$ for both using $P = V^2/R$ and then use $I = V/R$ to get a current of $0.3125\,\text{A}$. The power dissipated is then calculated using $P = I^2R$ and you get the above answers. However, doesn't that assume the voltage drop between the lightbulbs is $120\,\text{V}$ in both cases, and isn't that wrong?

I tried getting it another way where I said $P_1 = IV_1$, $P_2 = IV_2$, and $V_1+V_2=120\,\text{V}$. I solved the voltage drop on the $60\,\text{W}$ lightbulb to be $45\,\text{V}$ and $75\,\text{V}$ on the $100\,\text{W}$ one. Then, current is solved to be $4/3 \,\text{A}$ which lets us solve the resistance for each one as $33.75\,\Omega$ and $56.25\,\Omega$. Then using the formula $P = V^2/R$, the original wattages are found as the answer. Why is it right to assume $120\,\text{V}$ for both bulbs?

Answer 1

Why is the power dissipated not simply the wattages of the bulbs?

The power rating of a bulb is calculated assuming that the bulb will be used in a normal lighting circuit, that is, it will be in a parallel circuit, receiving the full domestic supply voltage, which I assume is 120V in your country.

A 60W bulb doesn't "know" it's supposed to pull 60W. It has a (fairly) constant resistance (once it has warmed up) which determines how much current it will draw when provided with a given voltage.

So for your series circuit you need to use 120V to calculate each bulb's resistance in ohms from its power rating. Then you can work out the total series resistance, and hence the total current for the curcuit, as you have done.

Answer 2

Answering the question from the title directly:

The bulb with lower power rating has a higher resistance (at operating temperature, but we have to assume it to be approximately constant), because at nominal voltage, less current must flow through it compared to the bulb with higher rating.

In a series circuit, the current is equal at all points.

The following assumes constant $R$ for each bulb.

The voltage drop at each bulb can be calculated using $V = RI$ (assuming constant $R$ despite changing temperature of the filament). Higher resistance leads to higer voltage drop at the bulb with lower wattage.

For 60W @ 120V we need 0.5A of current and with $R=V/I$ we get $R_{60W} = 240\Omega$. The other bulb has $R_{100W}= 144\Omega$.

Power is calculated using $P=VI$, and since $I$ is constant for both bulbs, higher resistance leads to higer power, more brightness.

The current is $I=V / (R_{60W} + R_{100W}) = 120V / 384 \Omega$

Answer 3

A 60W bulb means a bulb that when connected to it's rated supply voltage draws 60W. Unless otherwise stated one assumes that the rated supply voltage is the normal supply voltage in your country.

Your teacher is assuming that the bulbs act like resistors. If we make that assumption then we can solve your question in two steps.

1. Use the power ratings of the bulb and it's normal operating voltage to calculate the resistances.
2. Use the resistances calculated in the first step to calculate the behaviour of your actual circuit.

The first part of the question can be answered without actually doing the calculation. The lower rated bulb will have a higher resistance which means it will take a larger proportion of the voltage in the series circuit which means it will receive more power. For the latter part you have to actually work out the sums.

In practice the teachers assumption is highly inaccurate, but like with many textbook problems we have to run with it because we haven't been given any better information. In reality the resistance of an incandescent bulb increases singificantly with temperature. When you first turn a bulb on it draws far more current than it does after warming up.

What does this mean for our problem? it means two things.

1. The total power delivered to the two bulbs will be higher than the nieve calculations would suggest because the bulbs are cooler than they would be in normal operation.
2. The ratio of power delivered will be more extreme than the nieve calculations would suggest because the higher resistance (lower power rating) bulb will heat up more than the lower resistance (higher power rating) bulb.

Answer 4

One first assume that the resistance of the bulbs does no depend on the voltage/current/temperature.

The power of a bulb $P= \frac {V^2}{R}$ where $V$ is the potential difference across the bulb and $R$ is the resistance of the bulb.

For the same working voltage $V$ because $P_{\rm 100W}> P_{\rm 60W}$ then $R_{\rm 100W}< R_{\rm 60W}$.

The power $P=I^2R$ where $I$ is the current and for bulbs connected in series because the current is the same through both bulbs the bulb with the larger resistance (60W bulb) will dissipate more power ie be brighter, than the bulb with the smaller resistance (100W bulb).

Answer 5

The tungsten incandescent bulb with the highest resistance will have the greatest voltage drop and be the brightest since it will consume the most power ($RI^2$). This is the $60 \,\text{W}$ bulb as previously noted by other authors.

However the other authors are wrong when they calculate voltage, $V$, current, $I$, and power, $P$. They assume incorrectly that the incandescent bulb is ohmic, $V/I=R$, resistance is constant. They use the equation $P=V^2/R$ and assume $R$ is constant with $V$, which is wrong. They must use $P \propto V^{1.54}$ and $I \propto V^{0.54}$.

If we want to find the voltage through each incandescent bulb and the current we need to make some assumptions. The $60\,\text{W}$, and $100\,\text{W}$, are all rated at $120 \,\text{V}$. The current at $120 \,\text{V}$ through each is given by $P/V$, or $0.50\,\text{A}$ and $0.833\,\text{A}$ respectively. The current for all bulbs in series at $120 \,\text{V}$ must be the same for all. The current though the circuit is given by the following formulae: $$I=0.5\,\text{A} \times\left(\frac{x}{120\,\text{V}}\right)^{0.54} =0.833\,\text{A}\times\left(\frac{y}{120\,\text{V}}\right)^{0.54} \\ x + y=120 \,\text{V}$$ where $x$ is the voltage across the $60 \,\text{W}$ bulb and $y$ is the voltage across the $100 \,\text{W}$ bulb.

If we solve these equations simultaneously, we find that $x=86.45 \,\text{V}$, $y=33.55 \,\text{V}$ and the current, $= 0.418\,\text{A}$. The respective wattage is $VI$ or $36.18\,\text{W}$, $14.02\,\text{W}$. The respective resistance is $V/I$ or $207 \,\Omega$, $80.2 \,\Omega$. So the $60 \,\text{W}$ has the greatest resistance and will be the brightest. We see the power calculated above agrees with the $100 \,\text{W}$ bulb “answer” but not the $60 \,\text{W}$ bulb “answer” unless the 3 and 2 are transposed.

Answer 6

Normally lightbulbs are wired in parallel so that they all get the same voltage. In this configuration, the lightbulb with the highest resistance will dissipate the least amount of power ($V^2/R$). Putting two lightbulbs in series is non-standard. In such a configuration, the current is common and the voltage drops are different. Thus the power is given by $I^2R$, and the lightbulb with the higher resistance will dissipate the most power.

Details

A $60 \,\text{W}$ lightbulb has a resistance of $240 \,\Omega$. You can get this value by solving the for $R$ in the equation $P = V^2/R$, where $P$ is the nominal power of $60 \,\text{W}$, and $V$ is the nominal voltage of $120 \,\text{V}$. And, using the same approach, you can find the resistance for the $100 \,\text{W}$ light bulb, which is $144 \,\Omega$. As we expect, the lower-wattage lightbulb has a higher resistance. The total resistance of the circuit, when you put the two light bulbs in series is $R_{60} + R_{100} = 384 \,\Omega$ (I'm assuming no internal impedance of the power supply). The power rating on the bulbs when wired in series will not be the nominal powers of $60\,\text{W}$ and $100 \,\text{W}$ but will be a function of current and the total resistance of the circuit. The current in the circuit is given by $I = V/R_\text{total}$, which is $0.3125 \,\text{A}$. Next, if we assume that the brightness is a function of power only, then the power dissipated across the $60 \,\text{W}$ light bulb is $I^2R = 25.3906\,\text{W}$, and the power dissipated across the $100 \,\text{W}$ lightbulb is $14.0625 \,\text{W}$.