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When I connect two identical lightbulbs in series, how come they have equal brightness? Why can't one lightbulb have a larger voltage drop than the other? (i.e. the first lightbulb "uses up all the energy" and the second lightbulb is barely lit because "there isn't much energy left"?

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    $\begingroup$ Circuit elements in series have the same current, because every bit of charge that goes through one thing has to go through the other, and that's the definition of series. It's elements in parallel that have the same voltage drop. $\endgroup$ – Sean E. Lake Oct 12 '16 at 18:15
  • $\begingroup$ It's possible that in the first few seconds of current flow, the power output of the bulbs will be different. This is because the filament's resistance changes as they heat up. s $\endgroup$ – Howard Miller Oct 12 '16 at 20:19
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Resistors in series

When two resistors $R_1$ and $R_2$ are in series, the total resistance is:

$$R=R_1+R_2$$

If we apply a voltage $V$ across both resistors, the current $I$ is given by:

$$I=\frac{V}{R}=\frac{V}{R_1+R_2}$$

The voltage drop $V_i$ (for $i=1$ or $i=2$) across each resistor $R_i$ is:

$$V_i=R_iI=R_i\frac{V}{R_1+R_2}$$

So the voltage drops are not the same if $R_1\neq R_2$.

What determines the brightness of a bulb $i$ is its power output, $P_i$: $$P_i=V_iI=R_i\frac{V^2}{(R_1+R_2)^2}$$

So in series, the power of each bulb is proportional to each resistor $R_i$.

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  • $\begingroup$ $P_i$ is clearly not proportional to $R_i$ there. $\endgroup$ – Suzu Hirose Oct 12 '16 at 20:03
  • $\begingroup$ @SuzuHirose: strictly speaking you are right but if we look at $\Sigma R_i$ as the total resistance of the circuit, then $P_i\propto R_i$. It is in that spirit I formulated the conclusion. Thanks for your comment. $\endgroup$ – Gert Oct 12 '16 at 21:29
  • $\begingroup$ I can't believe that you defend this obvious error. $\endgroup$ – Suzu Hirose Oct 12 '16 at 22:48
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Why can't one lightbulb have a larger voltage drop than the other?

One can reason the correct answer without any math or knowledge of the bulb's I-V characteristics (other than to specify that the voltage across and current through a bulb are functionally related in some way, i.e., not independent).

(1) The lightbulbs are identical which implies that if they have the same current through, they have the same voltage across

(2) the lightbulbs are connected in series which means that all of the current through one bulb is through the other bulb, i.e., the bulbs have identical current through.

Thus, identical current through identical bulbs implies equal voltage across.

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  • $\begingroup$ Conceptually, why must the current be constant throughout? $\endgroup$ – yenny Oct 16 '16 at 22:20
  • $\begingroup$ To add another layer, my students often ask, "So when you add a second lightbulb, the current just KNOWS and will slow down immediately?" How does a current "know"? $\endgroup$ – yenny Oct 16 '16 at 22:20
  • $\begingroup$ @yenny, simply put, imagine that at some point in the circuit there is less current 'out' than 'in'. Necessarily, charge would accumulate at that point which would repel the flowing charge entering, reducing the current 'in', and repel the flowing charge leaving, increasing the current 'out' until the currents are equal. $\endgroup$ – Alfred Centauri Oct 16 '16 at 22:30
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Think of it in hydraulic or water like terms. It might be easier to visualize. You have a reservoir piped out to lets say ten similar devices downstream through one pipe. The first device the pipe comes to runs into a two way valve or y like valve. The water runs into the first device through an open valve and pressures up enough to open the other valve on the y to continue flowing to the next device. This only works if the device is there and functioning correctly or the second valve on the Y does not open. Then it comes to the next device and does the same thing. So far both devices are experiencing roughly the same pressure and thus operate with the same efficency. Just continue it on out to all ten devices with the same y valve set up on each. The pressure of the water is pretty close to the same at each device as it goes. Pressure equals current in this example. I realize it is not water and pressure we are dealing with but maybe this will help you understand.

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  • $\begingroup$ I thought this hydraulic example usually compared pressure to the VOLTAGE. This example does help. I'm curious then, about if the water first meets a wheel that has little resistance, and then after, a wheel that hast a lot of resistance. Wouldn't the current be inconsistent? The low-friction wheel has a high current and the high-friction wheel has a low current? Wouldn't that cause a blockage in the pipe? $\endgroup$ – yenny Oct 16 '16 at 22:25
  • $\begingroup$ A resistor is called that because it resists the flow of current. Electricity has to force its way through a resistor, and has to give up some energy to do it. The energy is turned into heat. Maybe only a little, maybe a lot. It depends on the value of the resistor and the amount of current. You could think of the resistor as a partial plug in a piece of pipe. Because the plug is there, the water current is reduced. It's the same with electricity. The electrical current is reduced. $\endgroup$ – william deets Oct 17 '16 at 2:55
  • $\begingroup$ So your device that needs more current just might not work as well but it is not a resistor. The current will still be the same downstream. $\endgroup$ – william deets Oct 17 '16 at 2:55
  • $\begingroup$ And yes it usually is compared to voltage. I'm comparing it to pressure for this example only. Remember the "y "in the example. The device is effectively not blocking the pressure. $\endgroup$ – william deets Oct 17 '16 at 2:59
  • $\begingroup$ A resistor would be like a restricted orifice in the side of the "y" that continues the current or pressure in the example on. $\endgroup$ – william deets Oct 17 '16 at 3:11
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The simplest way to think about the program is to realize that all components in series must have the same current running through them. There's no where else for the current to go, so due to the idea of Conservation of Charge, an equal current must pass through each component (in series).

So, if we have two lightbulbs in series, they must see the same current passing through them. Then, realizing that each bulb is identical, we know that they must have the same resistance.

Finally, we apply the equation for power output through a given resistance (in this case the light bulbs), which is $P = I^2 R$. This equation comes from combining $P = I \cdot V$ and $V = IR$. Because each bulb has the same current and the same resistance, it must have the same power output, and therefore the same brightness.

Lastly, perhaps a more fundamental way to think about the problem is through circuit symmetry. If each lightbulb is identical, and they are the only two components in the circuit (either in series or parallel), then they must have the same brightness. In order for them to have different brightnesses, something about the lightbulbs must be different, be it in placement in the circuit or a physical property of the bulb itself. If your thought then is that a possible difference in placement might be which end is located nearest to the positive/negative terminal of the power supply, this doesn't end up factoring in (unless you're talking about the Hall effect, or the like). The direction of the current through a circuit is largely arbitrary. Depending on the metal you use to connect the circuit, the charge carriers might be negative (like in copper) or they could be positive (like in aluminum).

This kind of symmetry analysis can be used to solve much trickier resistance problems, such as the ever-popular Resistor Cube.

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