# Why do bulbs glow brighter when connected in parallel?

Consider a circuit powered by a battery. If light bulbs are attached in parallel, the current will be divided across all of them. But if the light bulbs are connected in series, the current will be the same in all of them. Then it looks like the bulbs should be brighter when connected in series, but actually, they are brighter when connected in parallel. Why is that?

The bulbs will only appear brighter if the available current to the system is not limited. In that case the series bulbs will have a lower voltage across each individual bulb and they will appear dimmer. If the power input to the circuit is a constant than the total wattage output from all bulbs is also constant and the bulbs will all appear the same (assuming the filaments for the bulbs are all identical resistance).

In a typical simple circuit the power source will be a battery which attempts to hold a constant voltage across the circuit. In this case the voltage across the bulbs in parallel will be equal to the voltage of the battery and the current through the bulb will be defined by $V = IR$ where $R$ is the resistance of the filament. This means more current (and thus more power) will be drawn from a battery into the parallel circuit than a series one and the parallel circuit will appear brighter (but will drain your battery faster).

I crafted this answer for this question in the first place but since it got closed, I will post it here to at least contribute.

1) The brightness of a light bulb depends on various parameters, most of them being intrinsic properties of light bulbs. Essentially, the brightness depends upon the luminous flux of the light source. However, light sources which emit light with different wavelengths but same luminous flux can be perceived to have different brightness levels. Therefore, luminous flux is useful if we are comparing the brightness of light sources which emit light with same wavelength.

For incandescent light bulbs, brightness or luminous flux is directly related to the heat energy due to the flowing current in a conductor since these type of light bulbs are used by heating the filament until it emits visible light(assuming we have an incandescent light bulb here because other light sources like LED will have different properties). What is the term used to specify the heat energy generated by the flowing current per unit time? Power. Therefore, we should increase the power due to a current source as much as possible to increase the brightness of the light bulb. To find which parameters we should play with to increase the power, we can use Joule-Lenz law which states that: $$Q\propto I^2Rt$$ Therefore, since power is $\frac Wt$, we can derive the expression that is proportional to the power: $$P\propto I^2R$$ However, this expression can deceive you to think that increasing the resistance of the light bulb increases the brightness. Since altering the resistance will also decrease the current passing through the light bulb and even exponentially decrease the power, we can derive a more reliable formula by using the specialized form of Ohm's law($V=IR$). Assuming we have an ideal conductor here, one can find that; $$P\propto VI$$ Overall, you need to increase the emf of the current source to increase the brightness of the light bulb.

2) As all answerers pointed it out, when we wire light bulbs in parallel instead of in series, we decrease the equivalent resistance of the circuit; and therefore increase the current passing through the filaments of the light bulbs. This leads to more power each light bulb is getting(due to Joule-Lenz law) and brighter light bulbs.

Think about the power supplied to the bulb. Assume for a moment constant voltage source, and constant resistance for each bulb (not true for bulb but often used to simplify discussion at this level) then in series you have a total resistance of $2R$ and power $P= VI = \frac{V^2}{2R}$ . This power is split by two bulbs so each sees $V^2/4R$. When the bulbs are in parallel, each bulb sees the full voltage $V$ so $P=\frac{V^2}{R}$. Since a bulb glows brighter when it gets more power the ones in parallel will glow brighter.

See, the parallel combination of resistors reduces the effective resistance of the circuit. The voltage drop across each resistance is the same as the applied one. Hence most of the supplied voltage (the electrical energy) reaches the bulb. Hence it glows brighter. Now in the case of series combination of resistors, the effective resistance of the circuit increases. This in turn maximizes the voltage drop across the effective resistance as the voltage drop across each resistancee adds up. What energy remains only reaches the bulb. That's why the bulb glows less in series combination of resistors.

Let's first look at cases where this is not the case.
When the lamps are connected to a constant current source, current is indeed "divided" over the lamps in parallel. Assuming equal resistance R, the bulbs will both see a current of I/2, and the dissipated power in each is I²R/4 or in total I²R/2. In the case of the bulbs in series, the current I will flow through both bulbs, and the power consumed is I²R for each and 2I²R in total. Contrary to the claim in the question, the bulbs in series will burn brighter.

The same could be true when connecting them to a shunt wound DC generator. The bulbs in parallel have lower total equivalent resistance, pulling more current and lowering terminal voltage. Depending on the power rating of the generator and the bulbs, it could be that the bulbs in parallel represent a load too large for the generator, lowering the voltage over the shunt too far, which reduces the magnetic field, which causes the induced voltage to go down, lowering the magnetic field further etc.. The bulbs in series have higher total resistance and won't pull the voltage down so much. So in this case, they could again be the ones burning brightest.

When the bulbs are connected to a voltage source, the voltage is divided over the two bulbs in series, and the power consumed by each will be U²/4R. The bulbs in parallel each have a voltage of U over them, and therefor each consumes U²/R power. In this case the claim in the question is correct. the bulbs in parallel are brightest. This is the usual situation, voltage sources are much more common than current sources.

The first two examples assume that the bulbs in parallel and the bulbs in series weren't all (four) connected at the same time. If that was the case, those in parallel would always burn brighter.

Note: the assumption that incandescent light bulbs have constant resistance is completely wrong! Resistance changes with temperature:
For some metals, a linear function fits best: $R=R_0[1+\alpha(T-T_0)]$
For others, like tungsten, a power function fits better:

$\rho=0.06052*T^{1.203}$
with $\rho$ in $n\Omega.m$ T in Kelvin

A table of measured values can be found here At 2400°K, the resistance will be 14 times higher than at 273°K

All explanations given above are therefor only qualitatively correct (the result won't change, the same bulbs will be the brightest). An expression for the dependence of R on U or I can, if needed, be derived from the Stefan–Boltzmann law.

Take a battery of 4,5 V. Take one light bulb and some wires and arrange a simple circuit, measure the current, and lets say it is 2 A. Then, take another light bulb and connect it in parallel to the first one. Overall current will now be 4 A, it will increase. The rule is intact, as you can see, because in each branch now you have 2A which ads up to total of 4 A. It is because you added a new wire into the system, which is hooked on battery and electrons in that new wire are now on 4,5 potential difference, same as ones in the first wire. So they both produce the same current, 2 A, and these currents add up to 4. So adding more bulbs in parallel increases current and decreases R.

Lets start with two light bulbs, 120v AC, 60W each. If you connect them in parallel to the mains, each receives 120v, 0.5A (= 60W). If you now connect them in series, each light bulb now receives only 60v (due to voltage division), and assuming the same current (0.5A), it only receives 30W! Therefore, if the light bulbs light up at all, they each only has 30 watts of power, so they will be dimmer.