Why are lightbulbs with lower resistance brighter in parallel circuits, but lightbulbs with higher resistances are brighter in series circuits? When multiple lightbulbs are involved, why would a lightbulb with more resistance emit more light in a series circuit, but less in a parallel? (Pictures for reference)


 A: 
When multiple lightbulbs are involved, why would a lightbulb with more
resistance emit more light in a series circuit, but less in a
parallel? (Pictures for reference)

The current $I$ is the same through each bulb in a series circuit. So if we assume the light emitted is proportional to the power dissipated in its resistance, the light emitted from each bulb is proportional to
$$P=I^{2}R$$
for constant $I$ the greater the resistance $R$ of the bulb the greater the light emitted.
In a parallel combination of the same light bulbs, the voltage $V$ is the same across each bulb but the current is not. The power dissipated in each bulb is now proportional to
$$P=\frac{V^2}{R}$$
So for constant voltage $V$, the greater the bulb resistance the less the light emitted.
Hope this helps.
A: In the series circuit, each bulb has the same current but the bulb with most resistance receives most of the voltage, hence most power.
In the parallel circuit, each bulb has the same voltage, but the bulb with most resistance has the lowest current, hence least power.
Note that metal has a positive temperature coefficient, meaning that the resistance of the filament goes up when temperature goes up, so in the series circuit, the bulb with the most resistance will not only get more power than the other bulbs, but get even more power after heating up. This will further increase filament temperature, possibly leading to thermal runaway, leading to very fast lamp burn out, or at least a shortened life time. This is why it is a bad idea to connect bulbs in series, e.g. two bulbs of nominal 110 V, to a socket with 220 V. The illustration of the question, with three lamps in series with a 50.0 V supply, although beautifully drawn, is technically very silly. Also pointless to show FOUR labels with the current value, and to state resistance values at 3 decimals, where the filaments are assumed to be cold.
A: In the series circuit, all lightbulbs are run through by the same current. Given the current $I$ and resistance $R$, the total radiated power (over the whole radiation spectrum) of that single lightbulb is
$$P=R\cdot I^2$$
and thus, a higher resistance in the series is going to radiate more than the smaller resistances.
In the parallel circuit, all lightbulbs are exposed to the same voltage. Given voltage $U$ and resistance $R$, the total dissipated power of the single lightbulb is
$$P=\frac{U^2}{R}$$
and hence, the higher the resistance, the lower the power dissipation.
Note, however, that the total radiated power (over all lightbulbs) of a parallel circuit of lightbulbs is always more than the corresponding series circuit of the same lightbulbs radiates in total.
A: In the series circuit, a large part of the voltage is eaten by the resistances. In the parallel series a smaller part, so every bulb shines brighter.
If the lamps all have the same resistance, it will always be the case that in parallel they burn brighter because, with the same voltage, there will run more through each.
If the bulbs have different resistances, then in series the same current will run through all, but the voltages over them vary. The higher resistances have the highest voltages. Which means the currents through them produces most energy for high resistances.
In parallel, the voltage is the same for all but the current varies. Most current flows through the lowest resistance. And because the voltage is the same they will produce most light. The opposite for the series situation in which they have the same current but least voltage.
A: I detect the delicate aroma of homework. So I will not give you the answer, only give you several hints as to how to answer it.
What determines the amount of light a bulb emits? It will have something to do with voltage and current and resistance, those being the things that are available. But what do you think will be the quantity that corresponds to the amount of light emitted? The hint there is, suppose these are hot-filament bulbs where the bulb emits light from a hot chunk of wire. What quantity in electrical circuits might determine how hot that wire is? Or, more hintfully, how much heat does it produce per unit time?
Let us suppose you get a good answer to that. Call the quantity $X$.
For two bulbs, one with higher resistance than the other, and they are in series, what will be the value of X for each?  Say they have resistances $R$ and $r$, and $R = 2r$, and the voltage across the two bulbs in series is $V$. If it makes it easier, go ahead and use 30 volts for $V$, and 20 ohms and 10 ohms for the two resistances.
For the same two bulbs in parallel, and the same voltage applied, what will be the value of $X$ in each bulb?
If it is still not enough hints, when you buy light bulbs, what quantity is printed on them to indicate how bright they are?
A: This is basically the same as two other answers, but I prefer to deal with the stock equations directly.
Ohm's Law tells us that voltage (V) is equal to current (C) times resistance (R):
$V=IR$
Watt's Law tells us that power (P) is equal to voltage times current:
$P=VI$
You can do simple substitution from here to get the equations given in other answers, $P=I^2R$ and $P=\frac{V^2}{R}$. But I find it's not always helpful to say "we did algebra and these two things are equivalent and that's why this answer is true".
So let's look at the circuit more directly and hope it's helpful. :)
Series Circuit
Kirchhoff's Current Law tells us the current flowing into a node must equal the current flowing out. In the series circuit, each node only has an input and an output, so the input and output must be equal. Further, it follows that all the inputs and outputs must be equal because there's nowhere between node 1 and node 2 for the current to go.

Next, we have Kirchhoff's Voltage Law, that says for a closed loop the sum of voltages must be zero.

As discussed earlier, Ohm's Law tells us $V=IR$, so we can say that $V_1=I_1R_1$ and $V_2=I_2R_2$. Further, we know $I_1=I_2$ because of Kirchhoff's Current Law. This means $V_2=I_1R_2$, and $V_1=I_2R_1$.
We can re-arrange to get $I_1=\frac{V_1}{R_1}$ $=\frac{V_2}{R_2}$ $=I_2$. We're specifically interested in the fact that $\frac{V_1}{R_1}$ $=\frac{V_2}{R_2}$. If we know that one of the resistances is higher, the voltage on that side must also be higher for the equation to hold.
So whichever load has the highest resistance will also have the highest voltage drop: if $R_1>R_2$ then $V_1>V_2$.
Going back to Watt's Law, we can see that $P=VI$, so the power going through $R_1$ is given by $P_1=V_1I_1$, and the power through $R_2$ is similarly $P_2=V_2I_2$. Again, we know $I_1=I_2$, so we can re-arrange to get $I_1=\frac{P_1}{V_1}$ $=\frac{P_2}{V_2}$ $=I_2$.
As before, we're specifically interested in the fact that $\frac{P_1}{V_1}$ $=\frac{P_2}{V_2}$. If one voltage is higher, the power on that side must be higher for the equation to hold.
So whichever load has the most resistance has the highest voltage drop, and therefore the most power running through it: if $R_1>R_2$ then $V_1>V_2$ and therefore $P_1>P_2$.
Because a lightbulb's brightness is determined by the amount of power running through it, the bulb with the most resistance will be the brightest.
Parallel Circuit
With the parallel circuit case, Kirchhoff's laws still hold, but they're not particularly useful. The important thing to note is that the voltage going across each of the loads must be the same, because they're both tied to the same nodes.

Ohm's Law still holds, but now voltage is the constant across both loads. $V_1=I_1R_1$ and $V_2=I_2R_2$. Since $V_1=V_2$, this directly gives us $I_1R_1=I_2R_2$. We're multiplying instead of dividing, so we get the opposite effect from higher resistances as we did earlier: high resistance causes lower current.
That is, if $R_1>R_2$, then $I_1<I_2$. Watt's Law also still holds, so $P_1=V_1I_1$ and $P_2=V_2I_2$. Because $V_1=V_2$ we can re-arrange to get $V_1$ $=\frac{P_1}{I_1}$ $=\frac{P_2}{I_2}$ $=V_2$.
Again, we're interested in the fact that $\frac{P_1}{I_1}$ $=\frac{P_2}{I_2}$. This means whichever current is lower must have lower power for the equation to hold. Because higher resistance means lower current, it then follows that higher resistance means lower power: if $R_1>R_2$, then $I_1<I_2$, therefore $P_1<P_2$.
The lightbulb's brightness is still proportional to the power running through it, so the bulb with most resistance will be the dimmest.
Conclusion
In the series circuit, the high-resistance load has to dissipate more power than the low-resistance load, so it burns brighter.
In the parallel circuit, the power has an easier time getting through the low-resistance load, so it tends to bypass the high-resistance load, leading to less power running through the high-resistance load, causing it to be dimmer.
