Answer:

Okay, there is a slight inaccuracy in the problem. I said brightness is proportional to current. That is incorrect, it is proportional to POWER (voltage * current). But in this case voltage is constant so... details

Anyway:

So if the individual bulbs in string A and string B have equal brightness then:

Given:
V0 = A/C line voltage
i = current for that string
rA, rB = resistance of individual bulbs of type A or B

V0 * iA/100 = V0 * iB/50;

or

iA = 2iB

And since the V0 is the same (A/C wall voltage) we also have:

using v = i*r

iA*rA*100 = iB * rB * 50

substitute iA = 2iB from above:

2iB*rA*100 = iB*rB*50

or 4 rA = rB

For Q1:

Voltage across bulb A = V0 * rA/(rA + 49 * 4 * rA) = V0/197
Voltage across one of bulb B = V0 * 4 * rA/(rA + 49 * 4 * rA) = V0 * 4/197

Bulb B is brighter!

For Q2:

Voltage across bulb B = V0 * 4 * rA/(4 * rA + 49 * rA) = V0 * 4/53
Voltage across one of bulb A = V0 * rA/(4 * rA + 49 * rA) = V0 * 1/53

Bulb B is brighter! (And actually so bright it almost instantly burns out)

Not to be nitpicky, but in this case it is important, since you need to calculate voltage across individual bulbs which obv isn't constant to compare brightness, so that inaccuracy made the problem insolvable

Yup, we care about the voltage across each bulb, which means resistance is what we care about.

'A' bulbs are designed for 1.1V. `B` bulbs are designed for 2.2V. If the bulb gets less voltage than designed for they won't be as bright. If they get too much voltage then they will quickly (sometimes instantly burn out after being way too bright).

Therefore:

1) All bulbs are getting 2.2V. The 'B' bulbs are designed for this so they are normal while the 1 'A' bulb is designed to operate at 1.1V. So it will shine brighter and burn out.

2) Again, all bulbs are getting 2.2V. The 1 `B` bulb is designed for this, so it will be normal while all of the `A` bulbs are getting double there rated voltage and will burn out quickly. If they short as designed to keep others burning, then a cascade failure of the entire string will happen as the voltage goes up as each bulb fails.

And because they are rated by equal luminosity = equal power, which is proportional to the square of voltage (or current), the scalar for the resistance between the two is 4 not 2

if you walk in the desert for 40 days and 40 nights they will last 8 days

Are A bulbs not just generally brighter because they sustain the same brightness as B bulbs for 2x the amount of bulbs in the series?

No, because each bulb is the same brightness (not each string of bulbs).

Perhaps. But when I said voltage is constant, I meant for the individual situations of the all Bulb A string and all Bulb B string. so brightness is proportional to current in that specific derivative case.

In the two scenario's in the questions, the current is constant (as defined by each being a series of bulbs).

And you're still wrong...

?

About what?

Yes, thats all correct, with voltage and current being constant in these specific cases, but that makes your hint "brightness is proportional to current" still not helpful, if to solve the question you need to know that brightness is also proportional to voltage. I mean, it's not a problem when one googles brightness to check it, but as I said, I was in the shower thinking about this, so no google

If not wrong, at least barking up the wrong tree. We care about one bulb (because what does it mean to say the brightness of one hundred lamps its the same as fifty - that's a different question) and in that context brightness isn't proportional to current and voltage isn't constant.

resistance is futile

ja volt

lets not all get amped up here

Fyp