# Round bulb and Long bulb in Series Circuit

If you have a series circuit with a battery, one round bulb (with thicker filament), and one long bulb (with thinner filament), the round bulb does not light up and only the long bulb lights up.

Also, I understand that the voltage drop in the long bulb is greater, but I don't get how the charges distribute themselves like that. For example, if the round bulb comes first, how does the charges passing by the long bulb "know" to only give up a little bit of energy to save it up for the long bulb?

Why does the round bulb not light up?

-

Let's say that the round bulb has resistance $R_r$ and the long bulb has resistance $R_\ell$. Recall that the resistance $R$ of a resistor is related to it's length $\ell$, resistivity $\rho$, and cross sectional area $A$ by $$R = \frac{\ell\rho}{A}$$ Let's assume, for simplicity that the filaments of the two bulbs are made of the same material (so their resistivities are the same) and have the same length. Then this expression for the resistance tells us that $$R_r = \frac{A_\ell}{A_r}R_\ell$$ where $A_\ell$ is the cross sectional area of the filament in the long bulb, and $A_r$ is the cross sectional area of the filament in the round bulb. Notice that if the cross-sectional area of the long bulb is smaller (the filament is thinner) than that of the round bulb, then the resistance of the round bulb will be smaller.

Now, if both of these bulbs are in series with a battery , then Kirchoff's junction rule tells us that both bulbs have the same current $I$ going through them. Moreover, the power dissipated through Joule heating (which is what causes the filament to get hot and light up) in each resistor is given by $$P_\ell = I^2R_\ell, \qquad P_r = I^2 R_r$$ Using the relationship between the resistances derived above in terms of cross sectional areas, we therefore see that $$P_r = \frac{R_r}{R_\ell}P_\ell = \frac{A_\ell}{A_r}P_\ell$$ So the power dissipated through Joule heating in the round bulb is less that the power dissipated through Joule heating in the long bulb since the cross-sectional area of its filament in smaller.

This means, assuming that around the same fraction of the power is converted into energy in the form of visible light, that the round bulb will be dimmer than the long bulb.

In particular, you can make the round bulb arbitrarily dim compared to the long bulb by making its filament sufficiently thick.

-

As a contributor to, and user of, Matter & Interactions by Chabay and Sherwood (Wiley, 2010) (I coauthored the instructor's solution manual) I recognize this as a question from chapter 19 of the third edition. Specifically, see figure 19.51 of the text. I also now see that you have asked at least two questions straight out of Matter & Interactions. You need to be aware that the answers you have accepted thus far are not acceptable to instructors teaching the M&I story line correctly. The answers given by @joshphysics are mathematically correct, but they don't address the questions you are asking. Do you realize that? Joshphysics needs to know that the concept of resistance isn't introduced in M&I until chapter 20 and this isn't an acceptable concept in answering the questions you have asked. All that need be considered in how charged particles respond to electric fields and how to do a line integral along a path in an electric field to get a potential difference between two points.

Your question here is REALLY asking about how the surface charge distribution in a circuit with both a round bulb and a long bulb "knows" this or that. The simple answer is that it doesn't! What you need to do is to carefully re-read chapter 19, do the exercises AS YOU COME TO THEM IN THE TEXT, do the experiments AS YOU COME TO THEM IN THE TEXT, and then think really hard about what's being said. Of course, I won't divulge the answer here because homework solutions are considered off topic and they're not to be simply given to students in the first place.

I responded to your question about how putting two identical bulbs in parallel leads to an increased current in the circuit (before I realized what you're doing). You don't need the concept of resistance to answer that. I strongly suspect you aren't understanding chapter 19. Joshphysics's answer to that question won't make any sense to you until you get to chapter 20, at which time you will (hopefully) understand that resistance is a redundant concept. Furthermore, no M&I instructor would accept that answer as correct within the context of chapter 19.

-